Definite Integration

$ I=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\frac{12(3+[x])}{3+[\sin x]+[\cos x]}\right) d x $

Breakdown the Interval Since $\frac{\pi}{2} \approx 1.57$ and $-\frac{\pi}{2} \approx-1.57$, we break the integral at the integer points $x=-1,0,1$.

Notes on values :

• 1. For $x \in(0,1), 0<\cos x<1$, so $[\cos x]=0$. (At $x=0,[\cos x]=1$, but a single point doesn't change the integral).
  
  2. For $x \in\left[1, \frac{\pi}{2}\right), 0<\sin x<1$, so $[\sin x]=0$.

To find the total value, we integrate the constant values found above over their respective subintervals:

$ I=\int\limits_{-\pi / 2}^{-1} 6 d x+\int\limits_{-1}^1 12 d x+\int\limits_1^{\pi / 2} 16 d x $

$\Rightarrow $ $ I=6[x]_{-\pi / 2}^{-1}+12[x]_{-1}^1+16[x]_1^{\pi / 2} $

$\Rightarrow $ $ I=6\left(-1-\left(-\frac{\pi}{2}\right)\right)+12(1-(-1))+16\left(\frac{\pi}{2}-1\right) $

$\Rightarrow $ $ I=-6+3 \pi+24+8 \pi-16 \\ I=11 \pi+2 $

We are given the functional equation:

$ f\left(x^2+1\right)=x^4+5 x^2+2 $

Let $t=x^2+1$. This implies that $x^2=t-1$. Now, substitute this into the right side of the equation:

$ \begin{gathered} f(t)=(t-1)^2+5(t-1)+2 \\ f(t)=\left(t^2-2 t+1\right)+5 t-5+2 \\ f(t)=t^2+3 t-2 \end{gathered} $

Replacing $t$ back with $x$, the polynomial function is:

$ f(x)=x^2+3 x-2 $

Now, we calculate the definite integral from 0 to 3 :

$ \begin{gathered} \quad \int_0^3 f(x) d x=\int_0^3\left(x^2+3 x-2\right) d x \\ {\left[\frac{x^3}{3}+\frac{3 x^2}{2}-2 x\right]_0^3} \\ =\left(\frac{3^3}{3}+\frac{3\left(3^2\right)}{2}-2(3)\right)-(0) \\ =\left(\frac{27}{3}+\frac{27}{2}-6\right)=9+\frac{27}{2}-6=3+\frac{27}{2} \\ =\frac{6+27}{2}=\frac{33}{2} \end{gathered} $

$ \begin{aligned} &\begin{aligned} I= & \int_{\frac{\pi}{24}}^{\frac{5 \pi}{24}} \frac{1}{1+(\tan 2 x)^{1 / 3}} d x \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(1)\\ \Rightarrow & I=\int_{\frac{\pi}{24}}^{\frac{5 \pi}{24}} \frac{1}{1+\tan \left(2\left(\frac{5 \pi}{24}+\frac{\pi}{24}-4\right)\right)^{1 / 3}} d x \\ & I=\int_{\frac{\pi}{24}}^{\frac{5 \pi}{24}} \frac{1}{1+\tan \left(\frac{\pi}{2}-2 x\right)^{1 / 3}} d x \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(2)\end{aligned}\\ &\text { From equation (1) and (2) } \end{aligned} $

$ \begin{aligned} & 2 I=\int_{\frac{\pi}{24}}^{\frac{5 \pi}{24}} \frac{(\tan (2 x))^{1 / 3}+1}{1+(\tan 2 x)^{1 / 3}} d x \\ & =\left.(x)\right|_{\frac{\pi}{24}} ^{\frac{5 \pi}{24}} \Rightarrow I=\frac{1}{2} \times \frac{4 \pi}{24}=\frac{\pi}{12} \end{aligned} $

$ \begin{aligned} & \int_{-\frac{\pi}{2}}^{-1} \frac{1}{2} d x+\int_{-1}^0 \frac{1}{3} d x+\int_0^1 \frac{1}{4} d x+\int_1^{\frac{\pi}{2}} \frac{1}{5} d x \\ & =\frac{21 \pi-7}{60}=\frac{7}{60}(3 \pi-1) \end{aligned} $

$ \begin{aligned} &6 \int_1^x f(t) d t=3 x f(x)+x^2-4\\ &\text { By using leibnz rule }\\ &\begin{aligned} & 6 f(x)=3 f(x)+3 x f^{\prime}(x)+3 x^2 \\ & \frac{f(x)-x f^{\prime}(x)}{x^2}=1 \\ & \frac{d}{d x}\left(\frac{f(x)}{x}\right)=-1 f(x)=-x^2+k x \\ & 1=-1+k(\text { since } f(1)=1) \\ & f(x)=-x^2+2 x \\ & f(2)-f(3)=3 \end{aligned} \end{aligned} $

$ \begin{aligned} & \int_0^{\frac{\pi}{6}} \frac{\pi+4 x^{11}}{1-\sin \left(|x|+\frac{\pi}{6}\right)}+\frac{\pi-4 x^{11}}{1-\sin \left(|x|+\frac{\pi}{6}\right)} d x \\ & =2 \pi \int_0^{\frac{\pi}{6}} \frac{1}{1-\sin \left(x+\frac{\pi}{6}\right)} d x \\ & =2 \pi \int_0^{\frac{\pi}{6}} \frac{1+\sin \left(x+\frac{\pi}{6}\right)}{\cos ^2\left(x+\frac{\pi}{6}\right)} d x \\ & =2 \pi \int_0^{\frac{\pi}{6}} \sec ^2\left(x+\frac{\pi}{6}\right)+\tan \left(x+\frac{\pi}{6}\right) \sec \left(x+\frac{\pi}{6}\right) d x \\ & \Rightarrow 2 \pi\left[\tan \left(x+\frac{\pi}{6}\right)+\sec \left(x+\frac{\pi}{6}\right)\right]_0^{\frac{\pi}{6}} \\ & \Rightarrow 4 \pi \end{aligned} $

Let

$ I=\int_{0}^{2}\frac{1}{3^x+3}\,dx $

We will simplify the integrand first.

Since

$ 3^x+3=3(3^{x-1}+1), $

this form is not immediately very helpful. So let us use the substitution

$ 3^x=t $

Then,

$ t=3^x \quad \Rightarrow \quad \frac{dt}{dx}=3^x\ln 3=t\ln 3 $

So,

$ dx=\frac{dt}{t\ln 3} $

Now change the limits:

• When $x=0$, $t=3^0=1$

• When $x=2$, $t=3^2=9$

Thus,

$ I=\int_{1}^{9}\frac{1}{t+3}\cdot \frac{1}{t\ln 3}\,dt $

$ I=\frac{1}{\ln 3}\int_{1}^{9}\frac{1}{t(t+3)}\,dt $

Now use partial fractions:

$ \frac{1}{t(t+3)}=\frac{A}{t}+\frac{B}{t+3} $

So,

$ 1=A(t+3)+Bt $

$ 1=(A+B)t+3A $

Comparing coefficients:

$ A+B=0,\qquad 3A=1 $

Hence,

$ A=\frac{1}{3},\qquad B=-\frac{1}{3} $

Therefore,

$ \frac{1}{t(t+3)}=\frac{1}{3}\left(\frac{1}{t}-\frac{1}{t+3}\right) $

So,

$ I=\frac{1}{3\ln 3}\int_{1}^{9}\left(\frac{1}{t}-\frac{1}{t+3}\right)\,dt $

Now integrate:

$ I=\frac{1}{3\ln 3}\left[\ln t-\ln(t+3)\right]_{1}^{9} $

$ I=\frac{1}{3\ln 3}\left[\ln\frac{t}{t+3}\right]_{1}^{9} $

Substitute the limits:

$ I=\frac{1}{3\ln 3}\left(\ln\frac{9}{12}-\ln\frac{1}{4}\right) $

$ I=\frac{1}{3\ln 3}\left(\ln\frac{3}{4}-\ln\frac{1}{4}\right) $

$ I=\frac{1}{3\ln 3}\ln 3 $

$ I=\frac{1}{3} $

So, the value of the definite integral is

$ \boxed{\frac{1}{3}} $

Hence, the correct option is:

$ \boxed{\text{Option B}} $

$I=\int\limits_0^2\left[\frac{\sqrt{x\left(x^2+x+1\right)}}{(\sqrt{x+1}) \cdot\left(\sqrt{x^4+x^2+1}\right)}\right] d x$

$\begin{aligned} & =\int_0^2\left[\frac{\sqrt{x\left(x^2+x+1\right)}}{(\sqrt{x+1}) \cdot\left(\sqrt{x^4+2 x^2+1-x^2}\right)}\right] d x \\ & =\int_0^2\left[\frac{\sqrt{x\left(x^2+x+1\right)}}{(\sqrt{x+1}) \cdot\left(\sqrt{\left(x^2+1\right)^2-x^2}\right)}\right] d x \\ & =\int_0^2\left[\frac{\sqrt{x\left(x^2+x+1\right)}}{\left.(\sqrt{x+1}) \cdot\left\{\sqrt{\left(x^2+1+x\right)\left(x^2+1-x\right)}\right\}\right]}\right] d x \\ & =\int_0^2\left[\frac{(\sqrt{x}) \cdot\left(\sqrt{x^2+x+1}\right)}{(\sqrt{x+1}) \cdot\left(\sqrt{x^2+1+x}\right) \cdot\left(\sqrt{x^2+1-x}\right)}\right] d x \\ & =\int_0^2\left[\frac{\sqrt{x}}{(\sqrt{x+1}) \cdot\left(\sqrt{x^2-x+1}\right)}\right] d x \\ & =\int_0^2\left[\frac{\sqrt{x}}{\sqrt{(x+1)\left(x^2-x+1\right)}}\right] d x \\ & =\int_0^2\left[\frac{\sqrt{x}}{\sqrt{x^3+1}}\right] d x\end{aligned}$

Let $x=t^2 \Rightarrow d x=2 t d t$

$\boldsymbol{I}=\int_0^{\sqrt{2}}\left[\frac{2 t^2}{\sqrt{t^6+1}}\right] d t=\left(\frac{2}{3}\right) \int_0^{\sqrt{2}}\left[\frac{3 t^2}{\sqrt{t^6+1}}\right] d t$

Let $t^3=u \Rightarrow 3 t^2 d t=d u$

$\boldsymbol{I}=\left(\frac{2}{3}\right)^2 \cdot \int_0^{2 \sqrt{2}}\left[\frac{\boldsymbol{d u}}{\sqrt{u^2+1}}\right]$

Using standard integration formulaes :

$\int\left(\frac{1}{\sqrt{x^2+a^2}}\right) \boldsymbol{d x}=\ln \left|\boldsymbol{x}+\sqrt{\boldsymbol{x}^2+\boldsymbol{a}^2}\right|+C$

= $\left(\frac{2}{3}\right) \cdot\left[\ln \left|\boldsymbol{u}+\sqrt{\boldsymbol{u}^2+1}\right|\right]_0^{2 \sqrt{2}}$

$=\left(\frac{2}{3}\right) \cdot[\ln (2 \sqrt{2}+3)-\ln 1]=\left(\frac{2}{3}\right) \cdot[\ln (3+2 \sqrt{2})]$

The function $\boldsymbol{f}(\boldsymbol{x})$ is continuous at $\boldsymbol{x}=\frac{\boldsymbol{\pi}}{2}$ where

$\boldsymbol{f}(\boldsymbol{x})=\left\{\begin{array}{ll}\frac{1}{3} & ; x \leqslant \frac{\pi}{2} \\ \frac{b(1-\sin x)}{(\boldsymbol{\pi}-2 x)^2} & ; x>\frac{\pi}{2}\end{array} \cdots(i)\right.$.

We observe the LHL from equation $(i)$ :

$ \lim\limits_{x \rightarrow\left(\frac{\pi}{2}\right)^{-}} f(x)=f\left(\frac{\pi}{2}\right)=\frac{1}{3} \ldots \ldots(i i) $

We must have the RHL from equation (i) to be $\frac{1}{3}$ for the function $\boldsymbol{f}(\boldsymbol{x})$ to be continuous at $\boldsymbol{x}=\frac{\boldsymbol{\pi}}{2}$ :

$ \lim\limits_{\boldsymbol{x} \rightarrow\left(\frac{\boldsymbol{\pi}}{2}\right)^{+}} \boldsymbol{f ( x )}=\frac{1}{3} \Rightarrow \lim\limits _{\boldsymbol{x} \rightarrow\left(\frac{\boldsymbol{\pi}}{2}\right)^{+}} \frac{\boldsymbol{b}(1-\sin \boldsymbol{x})}{(\boldsymbol{\pi}-2 x)^2}=\frac{1}{3} \ldots \ldots(\boldsymbol{i i i}) $

$\Rightarrow $ $\lim\limits_{x \rightarrow\left(\frac{\boldsymbol{\pi}}{2}\right)^{+}} \frac{\boldsymbol{b}(1-\sin \boldsymbol{x})}{(\boldsymbol{\pi}-2 \boldsymbol{x})^2}=\frac{0}{0}$ format $\leftarrow$ Apply L'hôpital's rule

$\Rightarrow $ $\lim\limits_{x \rightarrow\left(\frac{\boldsymbol{\pi}}{2}\right)^{+}} \frac{b \cos \boldsymbol{x}}{4(\boldsymbol{\pi}-2 \boldsymbol{x})}=\frac{0}{0}$ format ค Apply L'hôpital's rule once more

$\Rightarrow $ $\lim\limits_{x \rightarrow\left(\frac{\pi}{2}\right)^{+}} \frac{b \sin x}{8}=\frac{b}{8} \ldots \ldots(i v)$

We can mathematically demonstrate conditional continuity of $f(x)$ at $x=\frac{\pi}{2}$ from equation (iii) and (iv) :

$ \frac{b}{8}=\frac{1}{3} \Rightarrow b=\frac{8}{3} \ldots \ldots(v) $

Now, I = $\int\limits_0^{3 \mathrm{~b}-6}\left|x^2+2 x-3\right| \mathrm{d} x$

$=\int\limits_0^{(3)\left(\frac{8}{3}\right)-6}\left|\boldsymbol{x}^2+2 \boldsymbol{x}-3\right| \boldsymbol{d} \boldsymbol{x}$

$=\int\limits_0^2\left|x^2+2 x-3\right| d x$

$=\int\limits_0^2|(x+3)(x-1)| d x$

$\left\{\begin{array}{l}(x+3)(x-1) \geqslant 0 ; x \in(-\infty,-3] \\ (x+3)(x-1) \leqslant 0 ; x \in[-3,1] \\ (x+3)(x-1) \geqslant 0 ; x \in[1, \infty)\end{array}\right.$

$=\int\limits_0^1(x+3)(1-x) d x$ $+\int\limits_1^2(x+3)(x-1) d x$

$=\int\limits_0^1\left(3-2 x-x^2\right) d x$ $+\int\limits_1^2\left(x^2+2 x-3\right) d x$

$\begin{aligned} & =\left[3 x-x^2-\frac{x^3}{3}\right]_0^1 +\left[\frac{x^3}{3}+x^2-3 x\right]_1^2\end{aligned}$

$\begin{aligned} & =\left(3-1-\frac{1}{3}\right) +\left(\frac{8}{3}+4-6\right)-\left(\frac{1}{3}+1-3\right)\end{aligned}$

= 4

Since $A$ is a singular matrix, its determinant must be zero.

We have

$ A=\begin{bmatrix} 1 & 3 & -1\\ 2 & 1 & \alpha\\ 0 & 1 & -1 \end{bmatrix} $

So,

$ |A|=0 $

Now let us calculate the determinant by expanding along the first row:

$ |A|=1\begin{vmatrix}1 & \alpha\\ 1 & -1\end{vmatrix} -3\begin{vmatrix}2 & \alpha\\ 0 & -1\end{vmatrix} +(-1)\begin{vmatrix}2 & 1\\ 0 & 1\end{vmatrix} $

$ =1[(1)(-1)-(\alpha)(1)]-3[(2)(-1)-(\alpha)(0)]-1[(2)(1)-(1)(0)] $

$ =(-1-\alpha)-3(-2)-2 $

$ =-1-\alpha+6-2 $

$ =3-\alpha $

Since $|A|=0$,

$ 3-\alpha=0 \Rightarrow \alpha=3 $

Now,

$ f(x)=\int_0^x (t^2+2t+3)\,dt $

Integrating,

$ f(x)=\left[\frac{t^3}{3}+t^2+3t\right]_0^x =\frac{x^3}{3}+x^2+3x $

We need maximum and minimum values of $f(x)$ in $[1,\alpha]=[1,3]$.

Now check whether $f(x)$ is increasing or decreasing.

$ f'(x)=x^2+2x+3 $

$ f'(x)=x^2+2x+3=(x+1)^2+2>0 \quad \text{for all } x $

So $f(x)$ is strictly increasing on $[1,3]$.

Hence,

• minimum value at $x=1$

• maximum value at $x=3$

Now,

$ m=f(1)=\frac{1}{3}+1+3=\frac{13}{3} $

and

$ M=f(3)=\frac{27}{3}+9+9=9+9+9=27 $

Therefore,

$ M-m=27-\frac{13}{3} =\frac{81-13}{3} =\frac{68}{3} $

So,

$ 3(M-m)=3\cdot \frac{68}{3}=68 $

Hence the correct answer is:

$ \boxed{68} $

So, the correct option is $\boxed{\text{Option B}}$.

$ I=\int_{-1}^{1}\left(\frac{x^3+|x|+1}{x^2+2|x|+1}\right)\,dx $

We will use the property of modulus carefully.

Since $|x|$ changes at $x=0$, split the integral into two parts:

$ I=\int_{-1}^{0}\left(\frac{x^3+|x|+1}{x^2+2|x|+1}\right)\,dx+\int_{0}^{1}\left(\frac{x^3+|x|+1}{x^2+2|x|+1}\right)\,dx $

For $-1 \le x < 0$

Here, $|x|=-x$.

So,

$ x^3+|x|+1=x^3-x+1 $

and

$ x^2+2|x|+1=x^2-2x+1=(x-1)^2 $

Thus,

$ I_1=\int_{-1}^{0}\frac{x^3-x+1}{(x-1)^2}\,dx $

For $0 \le x \le 1$

Here, $|x|=x$.

So,

$ x^3+|x|+1=x^3+x+1 $

and

$ x^2+2|x|+1=x^2+2x+1=(x+1)^2 $

Thus,

$ I_2=\int_{0}^{1}\frac{x^3+x+1}{(x+1)^2}\,dx $

Hence,

$ I=I_1+I_2 $

Now simplify both integrands.

Simplifying $I_1$

We divide:

$ \frac{x^3-x+1}{(x-1)^2}=\frac{x^3-x+1}{x^2-2x+1} $

On division,

$ \frac{x^3-x+1}{x^2-2x+1}=x+2+\frac{2x-1}{(x-1)^2} $

Also,

$ \frac{2x-1}{(x-1)^2}=\frac{2(x-1)+1}{(x-1)^2}=\frac{2}{x-1}+\frac{1}{(x-1)^2} $

So,

$ I_1=\int_{-1}^{0}\left(x+2+\frac{2}{x-1}+\frac{1}{(x-1)^2}\right)\,dx $

Integrating term by term,

$ \int x\,dx=\frac{x^2}{2}, \qquad \int 2\,dx=2x $

$ \int \frac{2}{x-1}\,dx=2\ln|x-1| $

$ \int \frac{1}{(x-1)^2}\,dx=\int (x-1)^{-2}\,dx=-\frac{1}{x-1} $

Therefore,

$ I_1=\left[\frac{x^2}{2}+2x+2\ln|x-1|-\frac{1}{x-1}\right]_{-1}^{0} $

At $x=0$:

$ \frac{0}{2}+0+2\ln 1-\frac{1}{-1}=1 $

At $x=-1$:

$ \frac{1}{2}-2+2\ln 2-\frac{1}{-2} =\frac{1}{2}-2+2\ln 2+\frac{1}{2} =-1+2\ln 2 $

So,

$ I_1=1-(-1+2\ln 2)=2-2\ln 2 $

Simplifying $I_2$

Now,

$ \frac{x^3+x+1}{(x+1)^2}=\frac{x^3+x+1}{x^2+2x+1} $

On division,

$ \frac{x^3+x+1}{(x+1)^2}=x-2+\frac{4x+3}{(x+1)^2} $

Also,

$ \frac{4x+3}{(x+1)^2}=\frac{4(x+1)-1}{(x+1)^2}=\frac{4}{x+1}-\frac{1}{(x+1)^2} $

So,

$ I_2=\int_{0}^{1}\left(x-2+\frac{4}{x+1}-\frac{1}{(x+1)^2}\right)\,dx $

Integrating term by term,

$ I_2=\left[\frac{x^2}{2}-2x+4\ln|x+1|+\frac{1}{x+1}\right]_{0}^{1} $

At $x=1$:

$ \frac{1}{2}-2+4\ln 2+\frac{1}{2}=-1+4\ln 2 $

At $x=0$:

$ 0-0+4\ln 1+1=1 $

So,

$ I_2=(-1+4\ln 2)-1=-2+4\ln 2 $

Adding $I_1$ and $I_2$

$ I=(2-2\ln 2)+(-2+4\ln 2)=2\ln 2 $

Therefore,

$ \boxed{\int_{-1}^{1}\left(\frac{x^3+|x|+1}{x^2+2|x|+1}\right)\,dx=2\log_e 2} $

So, the correct option is:

$ \boxed{\text{Option B}} $

Let

$ I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{32\cos^4 x}{1+e^{\sin x}}\,dx $

We will use the standard property:

If

$ I=\int_{-a}^{a} f(x)\,dx, $

then

$ I=\int_{-a}^{a} f(-x)\,dx $

and hence

$ 2I=\int_{-a}^{a} \bigl(f(x)+f(-x)\bigr)\,dx $

Now here,

$ f(x)=\frac{32\cos^4 x}{1+e^{\sin x}} $

So,

$ f(-x)=\frac{32\cos^4(-x)}{1+e^{\sin(-x)}} $

Using $\cos(-x)=\cos x$ and $\sin(-x)=-\sin x$,

$ f(-x)=\frac{32\cos^4 x}{1+e^{-\sin x}} $

Therefore,

$ f(x)+f(-x) = 32\cos^4 x\left(\frac{1}{1+e^{\sin x}}+\frac{1}{1+e^{-\sin x}}\right) $

Now simplify the bracket:

$ \frac{1}{1+e^{\sin x}}+\frac{1}{1+e^{-\sin x}} $

Let $t=e^{\sin x}$. Then $e^{-\sin x}=\frac{1}{t}$. So,

$ \frac{1}{1+t}+\frac{1}{1+\frac{1}{t}} = \frac{1}{1+t}+\frac{t}{1+t} = 1 $

Hence,

$ f(x)+f(-x)=32\cos^4 x $

So,

$ 2I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} 32\cos^4 x\,dx $

Thus,

$ I=16\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \cos^4 x\,dx $

Since $\cos^4 x$ is an even function,

$ I=32\int_{0}^{\frac{\pi}{4}} \cos^4 x\,dx $

Now use the identity

$ \cos^4 x=\left(\cos^2 x\right)^2 =\left(\frac{1+\cos 2x}{2}\right)^2 $

which gives

$ \cos^4 x=\frac{3+4\cos 2x+\cos 4x}{8} $

Therefore,

$ I=32\int_0^{\frac{\pi}{4}} \frac{3+4\cos 2x+\cos 4x}{8}\,dx $

$ I=4\int_0^{\frac{\pi}{4}} (3+4\cos 2x+\cos 4x)\,dx $

Now integrate termwise:

$ \int 3\,dx=3x,\qquad \int 4\cos 2x\,dx=2\sin 2x,\qquad \int \cos 4x\,dx=\frac{\sin 4x}{4} $

So,

$ I=4\left[3x+2\sin 2x+\frac{\sin 4x}{4}\right]_0^{\frac{\pi}{4}} $

At $x=\frac{\pi}{4}$,

$ 3x=\frac{3\pi}{4},\qquad \sin 2x=\sin \frac{\pi}{2}=1,\qquad \sin 4x=\sin \pi=0 $

At $x=0$,

$ 3x=0,\qquad \sin 0=0 $

Hence,

$ I=4\left(\frac{3\pi}{4}+2\right) $

$ I=3\pi+8 $

So the correct answer is:

$ \boxed{3\pi+8} $

Hence, Option B is correct.

$\begin{aligned} & f(a)=\frac{1}{2}\left(2\left(2^a+2^{-a}\right)+\left(3^a+3^{-a}\right)\right) \\ & a=\frac{1}{2}(2 \times 2+2)=3 \\ & I=\int_{\ln (3-1)}^{\ln 3} \frac{e^{2 x}}{e^{4 x}-1} d x \\ & \text { put } e^{2 x}=t \\ & I=\frac{1}{2} \int_4^9 \frac{d t}{t^2-1} \\ & I=\frac{1}{4}\left(\ln \left(\frac{t-1}{t+1}\right)\right)_4^9 \\ & I=\frac{1}{4} \ln \left(\frac{4}{3}\right)\end{aligned}$

Let

$ I=\int_0^{\infty}\frac{\ln x}{x^2+4}\,dx $

We will use a standard substitution, as done in NCERT-type definite integral problems.

Take

$ x=2t $

Then

$ dx=2\,dt $

and when $x=0,\ t=0$, and when $x\to\infty,\ t\to\infty$.

So,

$ I=\int_0^{\infty}\frac{\ln(2t)}{(2t)^2+4}\cdot 2\,dt $

Simplify the denominator:

$ (2t)^2+4=4t^2+4=4(1+t^2) $

Hence,

$ I=\int_0^{\infty}\frac{\ln(2t)}{4(1+t^2)}\cdot 2\,dt =\frac12\int_0^{\infty}\frac{\ln(2t)}{1+t^2}\,dt $

Now use

$ \ln(2t)=\ln 2+\ln t $

So,

$ I=\frac12\int_0^{\infty}\frac{\ln 2}{1+t^2}\,dt+\frac12\int_0^{\infty}\frac{\ln t}{1+t^2}\,dt $

That is,

$ I=\frac{\ln 2}{2}\int_0^{\infty}\frac{dt}{1+t^2}+\frac12\int_0^{\infty}\frac{\ln t}{1+t^2}\,dt $

Now,

$ \int_0^{\infty}\frac{dt}{1+t^2}=\left[\tan^{-1} t\right]_0^{\infty}=\frac{\pi}{2} $

Also, let

$ J=\int_0^{\infty}\frac{\ln t}{1+t^2}\,dt $

Using the substitution $t=\frac1u$, we get

$ dt=-\frac{1}{u^2}\,du $

Then

$ J=\int_{\infty}^{0}\frac{\ln(1/u)}{1+1/u^2}\left(-\frac{1}{u^2}\right)du $

This simplifies to

$ J=\int_0^{\infty}\frac{-\ln u}{1+u^2}\,du =-J $

Therefore,

$ J=0 $

So,

$ I=\frac{\ln 2}{2}\cdot \frac{\pi}{2} =\frac{\pi\ln 2}{4} $

Hence, the value of the integral is

$ \boxed{\frac{\pi\ln 2}{4}} $

So, the correct option is:

$ \boxed{\text{Option B}} $

$\begin{aligned} & \text { Put } x=2 t \Rightarrow d x=2 d t \\ & I=\int_0^{\infty} \frac{\ell n 2 t}{4 t^2+4}(2 d t)=\frac{1}{2} \int_0^{\infty} \frac{\ell n 2+\ell n t}{t^2+1} d t \\ & =\frac{1}{2} \int_0^{\infty} \frac{\ell n 2}{t^2+1} d t+\frac{1}{2} \int_0^{\infty} \frac{\ell n t}{t^2+1} d t\end{aligned}$

= $\left[\frac{\ln 2}{2} \tan ^{-1} \mathrm{t}\right]_0^{\infty}+\mathrm{I}_1$

$=\frac{\ell \mathrm{n} 2}{2} \cdot \frac{\pi}{2}+\mathrm{I}_1$

$\mathrm{I}_2=\frac{1}{2} \int_0^{\infty} \frac{\ell \mathrm{nt}}{\mathrm{t}^2+1} \mathrm{dt}$

$t=\frac{1}{u} \Rightarrow d t=-\frac{d u}{u^2}$

$\begin{aligned} & =\frac{1}{2} \int_{\infty}^0 \frac{\ell \mathrm{n}(1 / \mathrm{u})}{\frac{1}{\mathrm{u}^2}+1}\left(-\frac{\mathrm{du}}{\mathrm{u}^2}\right) \\ & \operatorname{Add} \Rightarrow \mathrm{I}_1=0 \\ & \mathrm{I}=\frac{\pi \ell \mathrm{n} 2}{4}\end{aligned}$

Since the coefficients $a,b,c$ are real, complex roots occur in conjugate pairs.

Given two roots are:

$ \alpha=1,\qquad \beta=1+i\sqrt{2} $

So the third root must be:

$ 1-i\sqrt{2} $

Hence the cubic polynomial is:

$ x^3+ax^2+bx+c=(x-1)\bigl(x-(1+i\sqrt{2})\bigr)\bigl(x-(1-i\sqrt{2})\bigr) $

Now first multiply the conjugate factors:

$ \bigl(x-(1+i\sqrt{2})\bigr)\bigl(x-(1-i\sqrt{2})\bigr) $

This is of the form:

$ (A-iB)(A+iB)=A^2+B^2 $

with

$ A=x-1,\qquad B=\sqrt{2} $

So,

$ \bigl(x-(1+i\sqrt{2})\bigr)\bigl(x-(1-i\sqrt{2})\bigr) =(x-1)^2+2 $

Therefore,

$ x^3+ax^2+bx+c=(x-1)\bigl((x-1)^2+2\bigr) $

Let $y=x-1$. Then:

$ (x-1)\bigl((x-1)^2+2\bigr)=(x-1)^3+2(x-1) $

Now expand:

$ (x-1)^3=x^3-3x^2+3x-1 $

So,

$ x^3+ax^2+bx+c=x^3-3x^2+3x-1+2x-2 $

$ =x^3-3x^2+5x-3 $

Now evaluate:

$ \int_{-1}^{1}(x^3+ax^2+bx+c)\,dx = \int_{-1}^{1}(x^3-3x^2+5x-3)\,dx $

Using symmetry:

• $x^3$ is an odd function, so

$ \int_{-1}^{1}x^3\,dx=0 $

• $5x$ is also odd, so

$ \int_{-1}^{1}5x\,dx=0 $

Thus,

$ \int_{-1}^{1}(x^3-3x^2+5x-3)\,dx = \int_{-1}^{1}(-3x^2-3)\,dx $

Now,

$ \int_{-1}^{1}x^2\,dx=\left[\frac{x^3}{3}\right]_{-1}^{1}=\frac13-\left(-\frac13\right)=\frac23 $

Therefore,

$ \int_{-1}^{1}(-3x^2)\,dx=-3\cdot \frac23=-2 $

Also,

$ \int_{-1}^{1}(-3)\,dx=-3(2)=-6 $

So total value:

$ -2-6=-8 $

Hence,

$ \int_{-1}^{1}\left(x^3+ax^2+bx+c\right)\,dx=-8 $

The correct option is:

$ \boxed{-8} $

So, Option C is correct.

Let

$ I=\int_0^1 \cot^{-1}(1+x+x^2)\,dx $

We will use the standard identity

$ \cot^{-1} t=\tan^{-1}\left(\frac{1}{t}\right)\quad \text{for } t>0 $

Since $1+x+x^2>0$ on $[0,1]$, we can write

$ I=\int_0^1 \tan^{-1}\left(\frac{1}{1+x+x^2}\right)\,dx $

Now use the formula

$ \tan^{-1} a-\tan^{-1} b=\tan^{-1}\left(\frac{a-b}{1+ab}\right) $

We try to express

$ \frac{1}{1+x+x^2} $

in a suitable form. Observe that

$ \tan^{-1}(x+1)-\tan^{-1}x = \tan^{-1}\left(\frac{(x+1)-x}{1+x(x+1)}\right) = \tan^{-1}\left(\frac{1}{1+x+x^2}\right) $

Hence,

$ \cot^{-1}(1+x+x^2)=\tan^{-1}(x+1)-\tan^{-1}x $

So,

$ I=\int_0^1 \left[\tan^{-1}(x+1)-\tan^{-1}x\right]dx $

Therefore,

$ I=\int_0^1 \tan^{-1}(x+1)\,dx-\int_0^1 \tan^{-1}x\,dx $

In the first integral, put $u=x+1$. Then when $x=0$, $u=1$, and when $x=1$, $u=2$. So

$ \int_0^1 \tan^{-1}(x+1)\,dx=\int_1^2 \tan^{-1}u\,du $

Thus,

$ I=\int_1^2 \tan^{-1}u\,du-\int_0^1 \tan^{-1}x\,dx = \int_1^2 \tan^{-1}x\,dx-\int_0^1 \tan^{-1}x\,dx $

Now use

$ \int \tan^{-1}x\,dx=x\tan^{-1}x-\frac{1}{2}\log(1+x^2) $

So,

$ \int_1^2 \tan^{-1}x\,dx = \left[x\tan^{-1}x-\frac{1}{2}\log(1+x^2)\right]_1^2 $

and

$ \int_0^1 \tan^{-1}x\,dx = \left[x\tan^{-1}x-\frac{1}{2}\log(1+x^2)\right]_0^1 $

Hence,

$ I= \left[x\tan^{-1}x-\frac{1}{2}\log(1+x^2)\right]_1^2 - \left[x\tan^{-1}x-\frac{1}{2}\log(1+x^2)\right]_0^1 $

This becomes

$ I= \left(2\tan^{-1}2-\frac{1}{2}\log 5\right) - \left(\tan^{-1}1-\frac{1}{2}\log 2\right) -\left(\tan^{-1}1-\frac{1}{2}\log 2\right) $

since at $x=0$, the value is $0$.

Now $\tan^{-1}1=\dfrac{\pi}{4}$, so

$ I=2\tan^{-1}2-\frac{1}{2}\log 5-2\left(\frac{\pi}{4}-\frac{1}{2}\log 2\right) $

$ I=2\tan^{-1}2-\frac{1}{2}\log 5-\frac{\pi}{2}+\log 2 $

Now combine the logarithms:

$ \log 2-\frac{1}{2}\log 5 = \frac{1}{2}\log 4-\frac{1}{2}\log 5 = \frac{1}{2}\log\left(\frac{4}{5}\right) = -\frac{1}{2}\log\left(\frac{5}{4}\right) $

Therefore,

$ I=2\tan^{-1}2-\frac{1}{2}\log\left(\frac{5}{4}\right)-\frac{\pi}{2} $

So the correct answer is

$ \boxed{2\tan^{-1}2-\frac{1}{2}\log_e\left(\frac{5}{4}\right)-\frac{\pi}{2}} $

Hence, the correct option is:

$ \boxed{\text{Option D}} $

Let the degree of the real polynomial $f$ be $n$.

Since $f(x)=f^{\prime}(x)f^{\prime\prime}(x)$ for all $x$, we compare degrees:

$ \begin{aligned} & \deg(f)=\deg(f^{\prime})+\deg(f^{\prime\prime}) \\ & n=(n-1)+(n-2)=2n-3 \\ & \Rightarrow n=3 \end{aligned} $

So $f$ must be a cubic polynomial. Write

$ \begin{aligned} & f(x)=a x^3+b x^2+c x+d \end{aligned} $

Given $f(0)=0$, we get $d=0$. Hence

$ \begin{aligned} & f(x)=a x^3+b x^2+c x \end{aligned} $

Now find derivatives:

$ \begin{aligned} & f^{\prime}(x)=3 a x^2+2 b x+c \\ & f^{\prime \prime}(x)=6 a x+2 b \end{aligned} $

Use the given condition $f(x)=f^{\prime}(x)f^{\prime\prime}(x)$:

$ \begin{aligned} & \left(a x^3+b x^2+c x\right)=\left(3 a x^2+2 b x+c\right)(6 a x+2 b) \end{aligned} $

Now compare coefficients of like powers of $x$.

For $x^3$:

$ \begin{aligned} & a=18 a^2 \\ & \Rightarrow a=\frac{1}{18}\ \ (\text{since }a\neq 0\text{ for a cubic}) \end{aligned} $

For $x^2$:

$ \begin{aligned} & b=6 a b+12 a b \\ & b=18ab \end{aligned} $

With $a=\frac{1}{18}$, this becomes $b=b$, so it gives no new condition on $b$.

For $x$:

$ \begin{aligned} & c=4 b^2+6 a c \\ & c=4b^2+\frac{c}{3} \quad (\because 6a=\frac{1}{3})\\ & \Rightarrow \frac{2c}{3}=4b^2 \Rightarrow c=6b^2 \end{aligned} $

For the constant term:

$ \begin{aligned} & 0=2bc \Rightarrow bc=0 \end{aligned} $

But $c=6b^2$, so $bc=b(6b^2)=6b^3=0 \Rightarrow b=0$, and then $c=0$.

So the polynomial is

$ \begin{aligned} & f(x)=\frac{1}{18}x^3 \end{aligned} $

Now compute the required value:

$ \begin{aligned} & f^{\prime}(x)=\frac{1}{6}x^2 \Rightarrow f^{\prime}(2)=\frac{1}{6}\cdot 4=\frac{2}{3} \\ & f^{\prime\prime}(x)=\frac{1}{3}x \Rightarrow f^{\prime\prime}(2)=\frac{1}{3}\cdot 2=\frac{2}{3} \\ & \int_0^2 f(x)\,dx=\int_0^2 \frac{1}{18}x^3\,dx=\frac{1}{18}\left[\frac{x^4}{4}\right]_0^2=\frac{1}{18}\cdot \frac{16}{4}=\frac{2}{9} \end{aligned} $

Add them:

$ \begin{aligned} & f^{\prime}(2)+f^{\prime\prime}(2)+\int_0^2 f(x)\,dx \\ &= \frac{2}{3}+\frac{2}{3}+\frac{2}{9}=\frac{14}{9} \end{aligned} $

Finally,

$ \begin{aligned} & 36\left(f^{\prime}(2)+f^{\prime\prime}(2)+\int_0^2 f(x)\,dx\right)=36\cdot \frac{14}{9}=56 \end{aligned} $

Let the given integral be denoted by $I$. We can split $I$ into two separate integrals:

$I = \int\limits_{-2}^2 (|\sin x| + [x \sin x]) dx = \int\limits_{-2}^2 |\sin x| dx + \int\limits_{-2}^2 [x \sin x] dx$

Let $I_1 = \int\limits_{-2}^2 |\sin x| dx$ and $I_2 = \int\limits_{-2}^2 [x \sin x] dx$.

Step 1: Evaluate $I_1$

Based on the property of definite integrals, if $f(x)$ is an even function (i.e., $f(-x) = f(x)$), then $\int_{-a}^a f(x) dx = 2 \int_0^a f(x) dx$.

Since $|\sin(-x)| = |-\sin x| = |\sin x|$, the function is even. Thus,

$I_1 = 2 \int\limits_0^2 |\sin x| dx$

In the interval $[0, 2]$, $2$ radians is less than $\pi \ (\approx 3.14)$, so $\sin x > 0$. Hence, $|\sin x| = \sin x$.

$I_1 = 2 \int\limits_0^2 \sin x dx$

$I_1 = 2 [-\cos x]_0^2 = 2 (-\cos 2 - (-\cos 0))$

$I_1 = 2 (1 - \cos 2) = 2 - 2 \cos 2$

Step 2: Evaluate $I_2$

Let $f(x) = x \sin x$. We check if it is even or odd:

$f(-x) = (-x) \sin(-x) = (-x) (-\sin x) = x \sin x = f(x)$.

Since $f(x)$ is an even function, $[x \sin x]$ is also an even function.

$I_2 = 2 \int\limits_0^2 [x \sin x] dx$

Now, let us analyze the behavior of $f(x) = x \sin x$ in the interval $x \in [0, 2]$:

• When $x = 0$, $f(0) = 0$.

• When $x = 2$, $f(2) = 2 \sin 2$.

Notice that $\pi/2 < 2 < 2\pi/3$, meaning $\sin 2$ is between $\sin(\pi/2)=1$ and $\sin(2\pi/3)=\frac{\sqrt{3}}{2} \approx 0.866$.

So, $1.732 < 2 \sin 2 < 2$. Hence, the maximum value of $f(x)$ at $x=2$ is between $1$ and $2$.

Let's verify if $f(x)$ is strictly increasing in $[0, 2]$. We differentiate $f(x)$ with respect to $x$:

$f'(x) = \sin x + x \cos x$

For $x \in (0, \pi/2]$, both $\sin x > 0$ and $\cos x \ge 0$, so $f'(x) > 0$.

For $x \in (\pi/2, 2]$, $\cos x < 0$. We can rewrite $f'(x)$ as:

$f'(x) = \cos x (\tan x + x)$

Since $\tan x$ is negative in the second quadrant and $2$ radians is smaller than $\tan^{-1}(-2)$, the expression $(\tan x + x)$ evaluates to a negative value. Therefore, $\cos x$ (negative) $\times$ $(\tan x + x)$ (negative) will yield $f'(x) > 0$.

Since $f(x)$ strictly increases from $0$ to a value between $1$ and $2$, there is exactly one real number $\alpha \in (0, 2)$ at which $f(\alpha) = 1$. So,

$\alpha \sin \alpha = 1$

For $x \in [0, \alpha)$, the function values are $0 \le x \sin x < 1$, which gives $[x \sin x] = 0$.

For $x \in [\alpha, 2]$, the function values are $1 \le x \sin x < 2$, which gives $[x \sin x] = 1$.

Now, split the integral $I_2$ at $x = \alpha$:

$I_2 = 2 \left( \int\limits_0^\alpha 0 dx + \int\limits_\alpha^2 1 dx \right)$

$I_2 = 2 [x]_\alpha^2 = 2(2 - \alpha) = 4 - 2\alpha$

Step 3: Finding the value of $\beta \sin \left(\frac{\beta}{2}\right)$

Combining $I_1$ and $I_2$, we get the total integral:

$I = I_1 + I_2 = (2 - 2 \cos 2) + (4 - 2\alpha) = 6 - 2 \cos 2 - 2\alpha$

$I = 2(3 - \cos 2) - 2\alpha$

According to the question, the total integral evaluates to:

$I = 2(3 - \cos 2) + \beta$

Comparing our result with the given equation, we find:

$\beta = -2\alpha$

Finally, substituting $\beta = -2\alpha$ into the expression $\beta \sin \left(\frac{\beta}{2}\right)$:

$\beta \sin \left(\frac{\beta}{2}\right) = -2\alpha \cdot \sin\left(\frac{-2\alpha}{2}\right)$

$\beta \sin \left(\frac{\beta}{2}\right) = -2\alpha \cdot \sin(-\alpha) = -2\alpha (-\sin \alpha)$

$\beta \sin \left(\frac{\beta}{2}\right) = 2 (\alpha \sin \alpha)$

From our earlier deduction in Step 2, we established that $\alpha \sin \alpha = 1$. Therefore:

$\beta \sin \left(\frac{\beta}{2}\right) = 2 (1) = 2$

Option B is the correct answer.

$I=\int\limits_{0}^{20\pi} \left(1-2\sin^2 x\cos^2 x\right)\cdot dx$

$=20\pi-\frac{1}{2}\int\limits_{0}^{20\pi} \sin^2 2x\,dx$

$= 20\pi-\frac{1}{2}(40)\int\limits_{0}^{\pi/2} \sin^2 2x$

$= 20\pi-\frac{20}{2}\left(x-\frac{\sin 2x}{2}\right)_{0}^{\pi/2}$

$= 20\pi-10\left(\frac{\pi}{2}\right)=15\pi$

$\int\limits_{0}^{1} \frac{e^x + e^{-x}}{1}\,dx + \int\limits_{1}^{2} \frac{e^x + e^{-x}}{1}\,dx + \int\limits_{2}^{3} \frac{e^x + e^{-x}}{2}\,dx$

= $\int\limits_{0}^{2} e^x\,dx + \int\limits_{0}^{2} e^{-x}\,dx + \frac{1}{2}\int\limits_{2}^{3} (e^x + e^{-x})\,dx$

= $\left(e^2 - 1\right) + \frac{e^{-2} - 1}{-1} + \frac{1}{2}\left((e^3 - e^2) - (e^{-3} - e^{-2})\right)$

= $\frac{e^3 + e^2}{2} + 1 - 1 - \frac{e^{-2}}{2} - \frac{e^{-3}}{2}$

= $\frac{e^3 + e^2 - e^{-2} - e^{-3}}{2}$

$\begin{aligned} &\text { Let, } \mathrm{I}=\pi^2 \int_{-1}^{3 / 2}|\mathrm{x} \sin \pi \mathrm{x}| \mathrm{dx}\\ &\begin{aligned} & =\pi^2\left\{\int_{-1}^1 \mathrm{x} \sin \pi \mathrm{xdx}-\int_1^{3 / 2} \mathrm{x} \sin \pi \mathrm{xdx}\right\} \\ & =\pi^2\left\{2 \int_0^1 \mathrm{x} \sin \pi \mathrm{xdx}-\int_{-1}^{3 / 2} \mathrm{x} \sin \pi \mathrm{xdx}\right\} \end{aligned} \end{aligned}$

$\begin{aligned} &\text { Consider }\\ &\begin{aligned} & \int \mathrm{x} \sin \pi \mathrm{xdx} \\ & -\mathrm{x} \cdot \frac{1}{\pi} \cos \pi \mathrm{x}+\int 1 \cdot \frac{1}{\pi} \cos \pi \mathrm{xdx} \\ & =--\frac{\mathrm{x}}{\pi} \cos \pi \mathrm{x}+\frac{\sin \pi \mathrm{x}}{\pi^2} \\ & \mathrm{I}=\pi^2\left\{2\left(-\frac{\mathrm{x}}{\pi} \cos \pi \mathrm{x}+\frac{\sin \pi \mathrm{x}}{\pi^2}\right)_0^1-\left(-\frac{\mathrm{x}}{\pi} \cos \pi \mathrm{x}+\frac{\sin \pi \mathrm{x}}{\pi^2}\right)_1^{3 / 2}\right\} \\ & =\pi^2\left\{\frac{2}{\pi}-\left(-\frac{1}{\pi^2}-\frac{1}{\pi}\right)\right\} \\ & =\pi^2\left\{\frac{3}{\pi}+\frac{1}{\pi^2}\right\} \\ & =3 \pi+1 \end{aligned} \end{aligned}$

$\begin{aligned} & I_1=\int_{-\frac{1}{2}}^1 2 x f(2 x(1-2 x)) d x \\ & \Rightarrow 2 x=t \Rightarrow 2 d x=d t \quad \Rightarrow I_1=\frac{1}{2} \int_{-1}^2 t f(t(1-t)) d t \\ & \Rightarrow 2 I_1=\int_{-1}^2(1-t) f(1-t)(1-(1-t)) d t \end{aligned}$

$ \Rightarrow 2{I_1} = \int\limits_{ - 1}^2 {f(t(1 - t)dt - \int\limits_{ - 1}^2 {tf(t(1 - t)dt} } $

$\begin{aligned} & \Rightarrow 2 \mathrm{I}_1=\mathrm{I}_2-2 \mathrm{I}_1 \\ & \Rightarrow 4 \mathrm{I}_1=\mathrm{I}_2 \\ & \Rightarrow \frac{\mathrm{I}_2}{\mathrm{I}_1}=4 \end{aligned}$

$\begin{aligned} & I=\int_0^\pi \frac{(x+3) \sin x}{1+3 \cos ^2 x} d x \quad\text{..... (i)}\\ & =\int_0^\pi \frac{(\pi-x+3) \sin (\pi-x)}{1+3 \cos ^2(\pi-x)} d x \\ & =\int_0^\pi \frac{(\pi+3) \sin x-x \sin x}{1+3 \cos ^2 x} d x \quad\text{..... (ii)} \end{aligned}$

Add (i) & (ii)

$2 I=\int_0^\pi \frac{(\pi+6) \sin x}{1+3 \cos ^2 x} d x$

Let $\cos x=t \Rightarrow-\sin x d x=d t$

$(\pi+6) \int_1^{-1} \frac{-d t}{\left(1+3 t^2\right)}=\left(\frac{\pi+6}{3}\right) \int_{-1}^1 \frac{d t}{t^2+\left(\frac{1}{\sqrt{3}}\right)^2}$

$\Rightarrow \quad 2 I=\left.\left(\frac{\pi+6}{3}\right) \cdot \frac{1}{\left(\frac{1}{\sqrt{3}}\right)} \cdot \tan ^{-1} \frac{t}{\frac{1}{\sqrt{3}}}\right|_{-1} ^1$

$\begin{aligned} & \Rightarrow \quad 2 I=\left(\frac{\pi+6}{\sqrt{3}}\right)\left[\tan ^{-1}(\sqrt{3})-\tan ^{-1}(-\sqrt{3})\right] \\ &=\left(\frac{\pi+6}{\sqrt{3}}\right) \cdot\left(\frac{\pi}{3}-\left(-\frac{\pi}{3}\right)\right) \\ &=\left(\frac{\pi+6}{\sqrt{3}}\right) \cdot \frac{2 \pi}{3} \\ & \Rightarrow \quad I=\frac{\pi(\pi+6)}{3 \sqrt{3}} \end{aligned}$

$\begin{aligned} & f(x)+2 f\left(\frac{1}{x}\right)=x^2+5 \\ & 2 f\left(\frac{1}{x}\right)+4 f(x)=2\left(\frac{1}{x^2}+5\right) \\ & 3 f(x)=\frac{2}{x^2}-x^2+5 \\ & f(x)=\frac{1}{3}\left(\frac{2}{x^2}-x^2+5\right) \\ & 2 g(x)-3 g\left(\frac{1}{x}\right)=x \\ & 2 g\left(\frac{1}{x}\right)-3 g(x)=\frac{1}{x} \\ & \text { Or } 4 g(x)-6 g\left(\frac{1}{x}\right)=2 x \\ & 6 g\left(\frac{1}{x}\right)-9 g(x)=\frac{3}{x} \\ & -5 g(x)=2 x+\frac{3}{x} \end{aligned}$

$ \begin{aligned} & \text { Or } g(x)=-\frac{1}{5}\left(2 x+\frac{3}{x}\right) \\ & \int_1^2 f(x) d x=\int_1^2 \frac{1}{3}\left(\frac{2}{x^2}-x^2+5\right) d x \end{aligned}$

$ \begin{aligned} &\begin{aligned} & =\frac{1}{3}\left[-\frac{2}{x}-\frac{x^3}{3}+5 x\right]_1^2 \\ & =\frac{1}{3}\left[\left(-\frac{2}{2}-\frac{8}{3}+10\right)-\left(-2-\frac{1}{3}+5\right)\right] \\ & =\frac{1}{3}\left[-1-\frac{8}{3}+10+2+\frac{1}{3}-5\right] \\ & \alpha=\frac{11}{9} \end{aligned}\\ &\text { Now, } 2 g(x)=x+3 g\left(\frac{1}{2}\right)\\ &\begin{aligned} & 2 g\left(\frac{1}{2}\right)=\frac{1}{2}+3 g\left(\frac{1}{2}\right) \\ & g\left(\frac{1}{2}\right)=-\frac{1}{2} \end{aligned} \end{aligned}$

$\begin{aligned} & \therefore \beta=\int_1^2 g(x) d x \\ & =\frac{1}{2} \int_1^2\left(x+3 g\left(\frac{1}{2}\right)\right) d x \\ & =\frac{1}{2}\left[\frac{x^2}{2}+3 g\left(\frac{1}{2}\right) x\right]_1^2 \\ & =0 \\ & \therefore 9 \alpha+\beta=11 \end{aligned}$

$\begin{aligned} & I=\int_{-1}^1 \frac{(1+\sqrt{|x|-x}) e^x+(\sqrt{|x|-x}) e^{-x}}{e^x+e^{-x}} d x \\ & =\int_0^1 \frac{(1+\sqrt{|x|-x}) e^x+(\sqrt{|x|-x}) e^{-x}}{e^x+e^{-x}} \end{aligned}$

$\begin{aligned} & +\left(\frac{(1+\sqrt{|x|+x}) e^{-x}+(\sqrt{|x|+x}) e^{-x}}{e^{-x}+e^x}\right) d x \\ = & \int_0^1 \frac{(1+\sqrt{|x|-x}+\sqrt{|x|+x})\left(e^x+e^{-x}\right)}{e^x+e^{-x}} d x \\ = & \int_0^1(1+\sqrt{|x|-x}+\sqrt{|x|+x}) d x \end{aligned}$

$\begin{aligned} & =\int_0^1(1+\sqrt{2 x}) d x=\left.x\right|_0 ^1+\left.\frac{\sqrt{2} x^{\frac{3}{2}}}{\frac{3}{2}}\right|_0 ^1 \\ & =1+\frac{2 \sqrt{2}}{3} \end{aligned}$

$\begin{aligned} & I=\int_0^\pi \frac{8 x}{4 \cos ^2 x+\sin ^2 x} d x \ldots(1) \\ & I=\int_0^\pi \frac{8(\pi-x)}{4 \cos ^2(\pi-x)+\sin ^2(\pi-x)} d x \\ & I=\int_0^\pi \frac{8(\pi-x)}{4 \cos ^2 x+\sin ^2 x} d x \ldots(2) \end{aligned}$

Adding (1) and (2)

$\begin{aligned} & 2 I=8 \pi \int_0^\pi \frac{1}{4 \cos ^2 x+\sin ^2 x} d x \\ & I=4 \pi \times 2 \int_0^\pi \frac{\sec ^2 x}{4 \tan ^2 x} d x \end{aligned}$

Put $\tan x=t$

$\sec ^2 x d x=d t$

$I=8 \pi \int_0^{\infty} \frac{d t}{4+t^2}$

$\begin{aligned} & I=8 \pi \frac{1}{2}\left(\tan ^{-1} \frac{t}{2}\right)_0^{\infty} \\ & I=4 \pi\left(\frac{\pi}{2}\right) \\ & I=2 \pi^2 \end{aligned}$

Step 1: Ensure the innermost function is greater than zero:

$ \log_6(3 + 4x - x^2) > 1 $

Step 2: Simplify the inequality from Step 1:

$ 3 + 4x - x^2 > 6 \implies x^2 - 4x + 3 < 0 $

Step 3: Solve the quadratic inequality:

$ (x - 1)(x - 3) < 0 $

This inequality indicates that $ x $ must lie between the roots, giving the interval $ x \in (1, 3) $.

With the domain of $ f(x) $ identified as $ (a, b) = (1, 3) $, we calculate the definite integral over $[0, b-a]$:

Calculate the Integral:

Given:

$ \int_0^{b-a} [x^2] \, dx $

For $ b - a = 2 $, we compute:

$ \int_1^2 [x^2] \, dx = \int_1^{\sqrt{2}} 1 \, dx + \int_{\sqrt{2}}^{\sqrt{3}} 2 \, dx + \int_{\sqrt{3}}^2 3 \, dx $

Computation of Each Integral Segment:

$\int_1^{\sqrt{2}} 1 \, dx = \sqrt{2} - 1$

$\int_{\sqrt{2}}^{\sqrt{3}} 2 \, dx = 2(\sqrt{3} - \sqrt{2})$

$\int_{\sqrt{3}}^2 3 \, dx = 3(2 - \sqrt{3})$

Summing these, we have:

$ 5 - \sqrt{2} - \sqrt{3} $

Conclusion:

The values for $ p, q, $ and $ r $ are $ p = 5 $, $ q = 2 $, and $ r = 3 $, with the greatest common divisor of these numbers being 1. Therefore, adding them together gives:

$ p + q + r = 5 + 2 + 3 = 10 $

To solve the problem, we start with the given equation:

$ 10 \int_1^x f(t) \, dt = 5x f(x) - x^5 - 9 $

By differentiating both sides with respect to $ x $, we have:

$ \frac{d}{dx}\left(10 \int_1^x f(t) \, dt\right) = \frac{d}{dx}(5x f(x) - x^5 - 9) $

The left side simplifies to $ 10 f(x) $. For the right side, using the product rule and the power rule, we get:

$ 10 f(x) = 5f(x) + 5x \frac{d}{dx} f(x) - 5x^4 $

Rearranging terms, we obtain:

$ 5 f(x) = 5x \frac{d}{dx} f(x) - 5x^4 $

Let $ y = f(x) $. Thus:

$ 5 y = 5x \frac{dy}{dx} - 5x^4 $

Dividing by 5, we have:

$ y = x \frac{dy}{dx} - x^4 $

Rewriting, we get:

$ \frac{dy}{dx} - \frac{y}{x} = x^3 $

This is a linear differential equation. The integrating factor (I.F.) is calculated as:

$ \text{I.F.} = e^{\int \frac{-1}{x} \, dx} = e^{-\ln x} = \frac{1}{x} $

Multiplying through by the integrating factor, we have:

$ y \cdot \frac{1}{x} = \int x^3 \cdot \frac{1}{x} \, dx = \int x^2 \, dx = \frac{x^3}{3} + C $

Thus, we solve for $ y $:

$ y = \frac{x^4}{3} + Cx $

Substituting back, we need to find $ C $ using the condition at $ x = 1 $:

Since $ \int_1^1 f(t) \, dt = 0 $, we substitute:

$ 10 \cdot 0 = 5 \cdot 1 \cdot f(1) - 1^5 - 9 \quad \Rightarrow \quad 0 = 5f(1) - 1 - 9 $

$ 5f(1) = 10 \quad \Rightarrow \quad f(1) = 2 $

Now use $ f(1) = 2 $ in the function:

$ 2 = \frac{1}{3} + C \cdot 1 \quad \Rightarrow \quad C = \frac{5}{3} $

The function $ f(x) $ is given by:

$ f(x) = \frac{x^4}{3} + \frac{5}{3}x $

To find $ f(3) $:

$ f(3) = \frac{3^4}{3} + \frac{5}{3} \cdot 3 = 27 + 5 = 32 $

Therefore, the value of $ f(3) $ is 32.

$\begin{aligned} & x^2=2 y \text { and } y-2 x-6=0 \\ & \frac{x^2}{2}-2 x-6=0 \\ & x^2-4 x-12=0 \\ & x^2-6 x+2 x-12=0 \\ & x(x-6)+2(x-6)=0 \\ & (x-6)(x+2)=0 \end{aligned}$

Point of intersection are $(6,18)$ and $(-2,2)$

$(-2,2)$ is in second quadrant

$\begin{aligned} & a=-2, b=2 \\ & I=\int_{-2}^2 \frac{9 x^2}{1+5^x} d x\quad\text{..... (i)} \end{aligned}$

$I=\int_{-2}^2 \frac{9 x^2}{1+5^{-x}} d x\quad\text{..... (ii)}$

$\begin{aligned} &\text { Adding (i) and (ii) }\\ &\begin{aligned} & 2 I=\int_{-2}^2 9 x^2 d x \\ & I=9 \int_0^2 x^2 d x \\ & I=9\left(\frac{x^3}{3}\right)_0^2 \Rightarrow I=24 \end{aligned} \end{aligned}$

$\begin{aligned} & I=4 \int_0^1 \frac{1}{\sqrt{3+x^2}+\sqrt{1+x^2}} d x \\ & =2 \int_0^1 \sqrt{3+x^2}-\sqrt{1+x^2} d x \end{aligned}$

$\begin{aligned} & =2\left[\int_0^1 \sqrt{3+x^2} d x-\int_0^1 \sqrt{1+x^2} d x\right] \\ & =2\left[\left(\frac{1}{2} x \sqrt{x^2+3}+\frac{3}{2} \ln \left|\sqrt{3+x^2}+x\right|\right)-\right. \\ & \left.\quad\left(\frac{1}{2} x \sqrt{1+x^2}+\frac{1}{2} \ln \left|\sqrt{1+x^2}+x\right|\right)\right]_0^1 \end{aligned}$

$\begin{aligned} & =2\left[\left(1+\frac{3}{2} \ln 3-\frac{3}{2} \ln \sqrt{3}\right)-\left(\frac{\sqrt{2}}{2}+\frac{1}{2} \ln (\sqrt{2}+1)\right)\right] \\ & =2\left(1+\frac{3}{4} \ln 3-\frac{1}{\sqrt{2}}-\frac{1}{2} \ln (\sqrt{2}+1)\right) \\ & =3 \ln \sqrt{3}+2-\sqrt{2}-\ln (\sqrt{2}+1) \\ & I-3 \ln \sqrt{3}=2-\sqrt{2}-\ln (\sqrt{2}+1) \end{aligned}$

$f^{\prime}(x)=x^3-9 x^2+20 x=x(x-4)(x-5)$

$\begin{aligned} & \therefore f(x)=\frac{x^4}{4}-\frac{9 x^3}{3}+\frac{20 x^2}{2} \\ & f(1)=\frac{1}{4}-3+10=\frac{29}{4}=\alpha \\ & \left.f(4)=\frac{256}{4}-3(64) \right\rvert\,+10(16)=32=\beta \\ & 4(\alpha+\beta)=4\left(\frac{29}{4}+32\right)=157 \end{aligned}$

$\begin{aligned} & I=80 \int_0^{\frac{\pi}{4}}\left(\frac{\sin \theta+\cos \theta}{9+16(2 \sin \theta \cdot \cos \theta)}\right) d \theta \\ & =80 \int_0^{\frac{\pi}{4}} \frac{\sin \theta+\cos \theta}{9-16(1-2 \sin \theta \cdot \cos \theta-1)} d \theta \end{aligned}$

$\begin{aligned} &=80 \int_0^{\frac{\pi}{4}} \frac{\sin \theta+\cos \theta}{9+16-16(\sin \theta-\cos \theta)^2} \mathrm{~d} \theta\\ &\text { Let } \sin \theta-\cos \theta=\mathrm{t}\\ &(\cos \theta+\sin \theta) \mathrm{d} \theta=\mathrm{dt}\\ &=80 \int_{-1}^0 \frac{\mathrm{dt}}{25-16 \mathrm{t}^2}\\ &=\frac{80}{16} \int_{-1}^0 \frac{\mathrm{dt}}{\left(\frac{5}{4}\right)^2-\mathrm{t}^2}\\ &\left.=\frac{5}{2\left(\frac{5}{4}\right)} \ln \left|\frac{\frac{5}{4}+t}{\frac{5}{4}-t}\right|\right]_{-1}^0\\ &=2 \ln (1)+4 \ln 3\\ &=4 \ln 3 \end{aligned}$

$\begin{aligned} &\begin{aligned} & \int_0^2 \mathrm{xF}^{\prime}(\mathrm{x}) \mathrm{dx}=6 \\ & =\left.\mathrm{xF}(\mathrm{x})\right|_0 ^2-\int_0^2 \mathrm{f}(\mathrm{x}) \mathrm{dx}=6 \\ & =2 \mathrm{~F}(2)-\int_0^2 \mathrm{xF}(\mathrm{x}) \mathrm{dx}=6[\therefore \mathrm{f}(2)=2 \mathrm{~F}(2)=2] \\ & \int_0^2 \mathrm{xF}(\mathrm{x}) \mathrm{dx}=-2 \quad \ldots(1) \\ & \Rightarrow \int_0^2 \mathrm{~F}(\mathrm{x}) \mathrm{dx}=-2\quad \ldots(2) \end{aligned}\\ &\text { Also }\\ &\begin{aligned} & \int_0^2 x^2 F^{\prime \prime}(x) d x=\left.x^2 F^{\prime}(x)\right|_0 ^2-2 \int_0^2 x F^{\prime}(x) d x=40 \\ & =4 F^{\prime}(2)-2 \times 6=40 \\ & F^{\prime}(2)=13 \\ & \therefore F^{\prime}(2)+\int_0^2 F(x)=13-2=11 \end{aligned} \end{aligned}$

$\begin{aligned} & g(x)=x 2+x^{\frac{7}{3}} \\ & g^{\prime}(x)=2 x+\frac{7}{3} x^{\frac{4}{3}} \\ & f(x)=\frac{g^{\prime}(x)}{x} \\ & f(x)=2+\frac{7}{3} x^{\frac{1}{3}} \\ & f\left(r^3\right)=2+\frac{7 r}{3} \\ & \sum_{r=1}^{15}\left(1+\frac{7}{3} r\right)=310 \end{aligned}$

$\int_\limits{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{96 x^2 \cos ^2 x}{\left(1+e^x\right)} d x \quad$ (Apply King Property)

$\begin{aligned} & \int_0^{\frac{\pi}{2}} 96 x^2 \cos ^2 x=48 \int_0^{\frac{\pi}{2}} x^2(1+\cos 2 x) d x \\ & 48\left[\left(\frac{x^3}{3}\right)_0^{\pi / 2}+\int_0^{\frac{\pi}{2}} x_1^2 \cos _{I I} 2 x d x\right] \\ & \Rightarrow \text { On solving } \pi\left(2 \pi^2-12\right) \\ & \Rightarrow \alpha=2, \beta=-12 \\ & \Rightarrow(\alpha+\beta)^2=100 \end{aligned}$

$\begin{aligned} & \mathrm{I}(\mathrm{~m}, \mathrm{~m})=\int_0^1 \mathrm{x}^{\mathrm{m}-1}(1-\mathrm{x})^{\mathrm{n}-1} \mathrm{dx} \\ & \text { Let } \mathrm{x}=\sin ^2 \theta \quad \mathrm{dx}=2 \sin \theta \cos \theta \mathrm{~d} \theta \\ & \mathrm{I}(\mathrm{~m}, \mathrm{n})=2 \int_0^{\pi / 2}(\sin \theta)^{2 \mathrm{~m}-1}(\cos \theta)^{2 \mathrm{n}-1} \mathrm{~d} \theta \\ & \mathrm{I}(9,14)+\mathrm{I}(10,13)=2 \int_0^{\pi / 2}(\sin \theta)^{17}(\cos \theta)^{27} \mathrm{~d} \theta \\ & +2 \int_0^{\pi / 2}(\sin \theta)^{19}(\cos \theta)^{25} \mathrm{~d} \theta \\ & =2 \int_0^{\pi / 2}(\sin \theta)^{17}(\cos \theta)^{25} \mathrm{~d} \theta \\ & =\mathrm{I}(9,13) \end{aligned}$

For I

Apply king ($\mathrm{P}-5$) and add

$\begin{aligned} & 2 I=\int_0^{\pi / 2} d x=\frac{\pi}{2} \Rightarrow I=\frac{\pi}{4} \\ & I_2=\int_0^{\pi / 2} \frac{x \sin x \cos x}{\sin ^4 x+\cos ^4 x} d x \end{aligned}$

Apply king and add

$\begin{aligned} & \mathrm{I}_2=\frac{\pi}{4} \int_0^{\pi / 2} \frac{\tan \mathrm{xsec}{ }^2 \mathrm{xdx}}{\tan ^4 \mathrm{x}+1} \\ & \text { put } \tan ^2 \mathrm{x}=\mathrm{t} \\ & \frac{\pi}{8} \int_0^{\infty} \frac{\mathrm{dt}}{\mathrm{t}^2+1} \\ & =\frac{\pi}{8} \cdot \frac{\pi}{2}=\frac{\pi^2}{16} \end{aligned}$

$\begin{aligned} &\text { Let } \ln \mathrm{x}=\mathrm{t} \Rightarrow \frac{\mathrm{dx}}{\mathrm{x}}=\mathrm{dt}\\ &\begin{aligned} & I=\int_2^4 \frac{e^{\frac{1}{1+t^2}}}{e^{\frac{1}{1+t^2}}+e^{\frac{1}{1+(6-t)^2}}} d t \\ & I=\int_2^4 \frac{\frac{1}{e^{1+(6-t)^2}}}{\frac{1}{e^{1+(6-t)^2}}+e^{\frac{1}{1+t^2}}} d t \\ & 2 I=\int_2^4 d t=(t)_2^4=4-2=2 \\ & I=1 \end{aligned} \end{aligned}$

$\begin{aligned} f(x) & =7 \tan ^8 x+7 \tan ^6 x-3 \tan ^4 x-3 \tan ^2 x \\ & =7 \tan ^6 x\left(1+\tan ^2 x\right)-3 \tan ^2 x\left(1+\tan ^2 x\right) \\ & =\left(7 \tan ^6 x-3 \tan ^2 x\right)\left(1+\tan ^2 x\right) \\ & =\left(7 \tan ^6 x-3 \tan ^2 x\right) \sec ^2 x\end{aligned}$

$\begin{aligned} I_1= & \int_0^{\frac{\pi}{4}} f(x) d x=\int_0^{\frac{\pi}{4}}\left(7 \tan ^6 x-3 \tan ^2 x\right) \sec ^2 x d x \\ & =\left.\left(\frac{7 \tan ^7 x}{7}-\frac{3 \tan ^3 x}{3}\right)\right|_0 ^{\frac{\pi}{4}}=1-1=0\end{aligned}$

$\begin{aligned} I_2= & \int_0^{\frac{\pi}{4}} x f(x) d x=\int_0^{\frac{\pi}{4}} x\left(7 \tan ^6 x-3 \tan ^2 x\right) \sec ^2 x d x \\ & =\left.x\left(\tan ^7 x-\tan ^3 x\right)\right|_0 ^{\frac{\pi}{4}}-\int_0^{\frac{\pi}{4}} 1 \cdot\left(\tan ^7 x-\tan ^3 x\right) d x \\ & =0-\int_0^{\frac{\pi}{4}} \tan ^3 x\left(\tan ^2 x-1\right)\left(\tan ^2 x+1\right) d x\end{aligned}$

$ \begin{aligned} & =\int_0^{\frac{\pi}{4}}\left(\tan ^3 x-\tan ^5 x\right) \sec ^2 x d x=\frac{\tan ^4 x}{4}-\left.\frac{\tan ^6 x}{6}\right|_0 ^{\frac{\pi}{4}} \\ & =\frac{1}{12} \end{aligned} $

Hence $7 I_1+12 I_2=1$

$ \begin{aligned} & \text { Let } \lim _{n \rightarrow \infty}\left(\frac{(n+1)(n+2) \ldots . .(2 n)}{n^n}\right)^{\frac{1}{n}}=y \\ & \Rightarrow \ln y=\lim _{n \rightarrow \infty} \frac{1}{n}\left(\sum_{r=1}^n \ln (n+r)-\sum_{r=1}^n \ln (n)\right) \\ & \Rightarrow \ln y=\lim _{n \rightarrow \infty} \frac{1}{n}\left(\sum_{r=1}^n \ln \left(1+\frac{r}{n}\right)\right) \\ & \Rightarrow \ln y=\int_0^1 \ln (1+x) d x \\ & \Rightarrow \quad y=e^0 \ln (1+x) d x \\ & \text { Let } P=\int_0^1 \ln (1+x) d x \end{aligned} $

$ \begin{aligned} &\begin{array}{ll} \Rightarrow & P=[x \cdot \ln (1+x)]_0^1-\int_0^1\left(\frac{x}{1+x}\right) d x \\ \Rightarrow & P=\ln 2-\int_0^1\left(\frac{x}{1+x}\right) d x \end{array}\\ &\text { Again, let } 1+x=t\\ &\begin{aligned} d x & =d t \\ & =\int_1^2\left(\frac{t-1}{t}\right) d t=\int_1^2\left(1-\frac{1}{t}\right) d t \\ & =[t-\ln t]_1^2 \\ & =2-\ln 2-1=1-\ln 2 \\ \therefore \quad P & =\ln 2-(1-\ln 2) \\ \Rightarrow \quad P & =\ln 2-1+\ln 2 \\ \Rightarrow \quad P & =2 \ln 2-1 \\ \Rightarrow \quad P & =\ln 4-\ln e \\ \Rightarrow \quad P & =\ln \frac{4}{e} \\ \Rightarrow \quad y & =e \ln \frac{4}{e} \\ y & =\frac{4}{e} \end{aligned} \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { Let } I_n=\int_0^{\frac{\pi}{2}} \tan ^n\left(\frac{x}{2}\right) d x \\ & I_{n-2}=\int_0^{\pi / 2}\left(\tan \frac{x}{2}\right)^{n-2} d x \\ & \Rightarrow \quad I_n+I_{n-2}=\int_0^{\frac{\pi}{2}}\left(\tan \frac{x}{2}\right)^{n-2} \times \sec ^2 \frac{x}{2} d x \\ & \Rightarrow \quad I_n+I_{n-2}=2 \int_0^1 t^{n-2} d t=\frac{2}{n-1} \end{aligned} \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad I_{14}+I_{12}=\frac{2}{13}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \\ & \Rightarrow \quad I_{12}+I_{10}=\frac{2}{11} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ & \Rightarrow \quad I_{10}+I_8=\frac{2}{9} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\\ & \Rightarrow \quad I_8+I_6=\frac{2}{7} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)\\ & \Rightarrow \quad I_6+I_4=\frac{2}{5} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(v)\\ & \Rightarrow \quad I_4+I_2=\frac{2}{3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(vi) \\ & \Rightarrow \quad I_2+I_0=2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(vii)\\ & \therefore(\mathrm{i})-(\mathrm{ii})+(\mathrm{iii})-(\mathrm{iv})+(\mathrm{v})-(\mathrm{vi})+(\mathrm{vii}) \\ & I_{14}=2\left[\sum_{n=1}^7 f(x)-\frac{\pi}{4}\right] \\ & \therefore f(n)=\frac{1}{13}-\frac{1}{11}+\frac{1}{9}-\frac{1}{7} \ldots . .+1 \\ & f(n)=\frac{(-1)^{n+1}}{2 n-1} \end{aligned} $

$ \text { } \begin{aligned} Let\,\,I & =\int_{-4}^5 \frac{d x}{\sqrt{20+x-x^2}} \\ I & =\int_{-4}^5 \frac{d x}{\sqrt{20-\left(x^2-x+\frac{1}{4}-\frac{1}{4}\right)}} \\ & =\int_{-4}^5 \frac{d x}{\sqrt{\frac{81}{4}-\left(x-\frac{1}{2}\right)^2}} \\ & =\left[\sin ^{-1}\left(\frac{2 x-1}{9}\right)+C\right]_{-4}^5 \end{aligned} $

$ =\sin ^{-1}\left(\frac{9}{9}\right)-\sin ^{-1}(-1)=\pi $

$ \begin{aligned} & \text { Let } I=\int_0^{\pi / 2} \frac{d x}{\cos x-\sqrt{3} \sin x} \\ & =\int_0^{\pi / 2} \frac{d x}{2\left(\cos x \cdot \frac{1}{2}-\frac{\sqrt{3}}{2} \sin x\right)} \\ & =\frac{1}{2} \int_0^{\pi / 2} \frac{d x}{\cos \left(x+\frac{\pi}{3}\right)} \\ & =\frac{1}{2} \int_0^{\pi / 2} \sec \left(x+\frac{\pi}{3}\right) d x \\ & =\frac{1}{2} \ln \left|\sec \left(x+\frac{\pi}{3}\right)+\tan \left(x+\frac{\pi}{3}\right)\right|_0^{\pi / 2}+C \\ & =\frac{1}{2} \ln \left|\sec \left(\frac{5 \pi}{6}\right)+\tan \left(\frac{5 \pi}{6}\right)\right| -\frac{1}{2} \ln \left|\sec \frac{\pi}{3}+\tan \frac{\pi}{3}\right| \\ & =\frac{1}{2} \ln (2 \sqrt{3}-3) \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { Let } I=\int_0^{\pi / 2} \sqrt{\tan x} d x \\ & \qquad \begin{aligned} I & =\int_0^{\pi / 2} \sqrt{\cot x} d x \\ \Rightarrow 2 I & =\int_0^{\pi / 2} \sqrt{2} \times \frac{\sin x+\cos x}{\sqrt{1-(\sin x-\cos x)^2}} d x \\ \Rightarrow I & =\frac{1}{\sqrt{2}} \int_0^{\pi / 2} \frac{d(\sin x-\cos x)}{\sqrt{1-(\sin x-\cos x)^2}} \\ & =\frac{1}{\sqrt{2}}\left[\sin ^{-1}(\sin x-\cos x)+C\right]_0^{\pi / 2} \\ & =\frac{1}{\sqrt{2}}\left(\sin ^{-1}(1)-\sin ^{-1}(-1)\right) \\ & =\frac{1}{\sqrt{2}} \times \frac{\pi}{2} \times 2=\frac{\pi}{\sqrt{2}} \end{aligned} \end{aligned} \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & I=\int_{-1}^1 \frac{\log 2-\log (1+x)}{\sqrt{1-x^2}} d x \\ & \Rightarrow I=\int_{-1}^1 \frac{\log 2}{\sqrt{1-x^2}} d x-\int_{-1}^1 \frac{\log (1+x)}{\sqrt{1-x^2}} d x \\ & \Rightarrow I=\left[\log 2 \sin ^{-1} x\right]_{-1}^1-\int \frac{\log (1+x)}{\sqrt{1-x^2}} d x \\ & =\pi \log 2-I_1 \\ & I_1=\int_{-1}^1 \frac{\log (1+x)}{\sqrt{1-x^2}} d x \end{aligned}\\ &\text { Put } \begin{aligned} x & =\sin \theta \\ d x & =\cos \theta d \theta \end{aligned} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \Rightarrow I_1=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{\log (1+\sin \theta)}{\cos \theta} \cos \theta d \theta \\ & \Rightarrow I_1=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \log (1+\sin \theta) d \theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned}\\ &\text { King's property, }\\ &\Rightarrow \quad I_1=\int_{\frac{-\pi}{2}}^{\pi / 2} \log (1-\sin \theta) d \theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

$ \begin{aligned} \Rightarrow \quad 2 I_1 & =\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \log \left(1-\sin ^2 \theta\right) d \theta \\ \Rightarrow \quad I_1 & =\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \log \cos \theta d \theta \\ & =2 \int_0^{\frac{\pi}{2}} \log \cos \theta d \theta \\ & =2\left(\frac{-\pi}{2} \log 2\right) \\ & =-\pi \log 2 \\ \therefore \quad I & =2 \pi \log 2 \end{aligned} $

$ \begin{aligned} I & =\int_0^{\frac{\pi}{4}} \frac{\sec x}{3 \cos x+4 \sin x} d x \\ & =\int_0^{\frac{\pi}{4}} \frac{\sec ^2 x}{3+4 \tan x} d x \end{aligned} $

Put $\tan x=t \Rightarrow \sec ^2 x d x=d t$

$ \begin{aligned} \Rightarrow \quad I & =\int_0^1 \frac{d t}{3+4 t} \\ & =\frac{1}{4}[\log 4 t+3]_0^1=\frac{1}{4} \log \frac{7}{3} \end{aligned} $

$ \text { } \begin{aligned} I & =\int_{-2}^4\left|\left(2-x^2\right)\right| d x \\ = & \int_{-2}^{-\sqrt{2}}\left(x^2-2\right) d x+\int_{-\sqrt{2}}^{+\sqrt{2}}\left(2-x^2\right) d x+\int_{\sqrt{2}}^{+4}\left(x^2-2\right) d x \\ = & {\left[\frac{x^3}{3}-2 x\right]_{-2}^{-\sqrt{2}}+\left[2 x-\frac{x^3}{3}\right]_{-\sqrt{2}}^{\sqrt{2}} } +\left[\frac{x^3}{3}-2 x\right]_{\sqrt{2}}^4 \end{aligned} $

$ \begin{aligned} & \begin{aligned} \Rightarrow & \left.\frac{-2 \sqrt{2}}{3}+2 \sqrt{2}\right)-\left(\frac{-8}{3}+4\right) \\ + & {\left[\left(2 \sqrt{2}-\frac{2 \sqrt{2}}{3}\right)-\left(-2 \sqrt{2}+\frac{2 \sqrt{2}}{3}\right)\right] } \\ + & {\left[\left(\frac{64}{3}-8\right)-\left(\frac{2 \sqrt{2}}{3}-2 \sqrt{2}\right)\right] } \end{aligned} \\ & \Rightarrow \frac{4 \sqrt{2}}{3}-\frac{4}{3}+4 \sqrt{2}-\frac{4 \sqrt{2}}{3}+\frac{40}{3}+\frac{4 \sqrt{2}}{3} \\ & =12+\frac{16 \sqrt{2}}{3} \end{aligned} $

$ \begin{aligned} &\begin{aligned} I & =\int_0^{\pi / 4} \frac{d x}{5 \cos ^2 x+16 \sin ^2 x+8 \sin x \cos x} \\ & =\int_0^{\pi / 4} \frac{\sec ^2 x d x}{5+16 \tan ^2 x+8 \tan x} \end{aligned}\\ &\text { Put, } \tan x=t \Rightarrow \sec ^2 x d x=d t\\ &\begin{aligned} I & =\int_0^1 \frac{d t}{16 t^2+8 t+5} \\ & =\frac{1}{16} \int_0^1 \frac{d t}{t^2+\frac{1}{2} t+\frac{5}{16}} \\ & =\frac{1}{16} \int_0^1 \frac{d t}{\left(t+\frac{1}{4}\right)^2+\frac{5}{16}-\frac{1}{16}} \\ & =\frac{1}{16} \int_0^1 \frac{d t}{\left(t+\frac{1}{4}\right)^2+\left(\frac{1}{2}\right)^2} \\ & =\frac{1}{16} \times \frac{1}{\frac{1}{2}} \tan ^{-1}\left(\frac{t+\frac{1}{4}}{\frac{1}{2}}\right)_0^1 \\ & =\frac{1}{8}\left[\tan ^{-1}\left(\frac{\frac{5}{2}}{2}\right)-\tan ^{-1}\left(\frac{1}{2}\right)\right] \\ & =\frac{1}{8} \tan ^{-1}\left(\frac{\frac{5}{2}-\frac{1}{2}}{1+\frac{5}{4}}\right)=\frac{1}{8} \tan ^{-1}\left(\frac{8}{9}\right) \end{aligned} \end{aligned} $

$I=\int_8^{18} \frac{1}{(x+2) \sqrt{x-3}} d x$

$ \begin{aligned}Put\,\,\, & x-3=t^2 \\ & d x=2 t d t \\ & I=\int_{\sqrt{5}}^{\sqrt{15}} \frac{2 d t}{\left(t^2+5\right) \cdot t}=2 \int_{\sqrt{5}}^{\sqrt{15}} \frac{d t}{t^2+5} \\ & =\frac{2}{\sqrt{5}}\left(\tan ^{-1} \frac{t}{\sqrt{5}}\right)_{\sqrt{5}}^{\sqrt{15}} \\ & =\frac{2}{\sqrt{5}}\left(\frac{\pi}{3}-\frac{\pi}{4}\right)=\frac{2}{\sqrt{5}}\left(\frac{\pi}{12}\right)=\frac{\pi}{6 \sqrt{5}} \end{aligned} $