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}} $

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

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

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

$ \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} $

$ \begin{aligned} I & =\int_1^2\left[x^2\right] d x \\ & =\int_1^{\sqrt{2}} 1 d x+\int_{\sqrt{2}}^{\sqrt{3}} 2 d x+\int_{\sqrt{3}}^2 3 d x \\ & =(\sqrt{2}-1)+2(\sqrt{3}-\sqrt{2})+3(2-\sqrt{3}) \\ & =\sqrt{2}-1+2 \sqrt{3}-2 \sqrt{2}+6-3 \sqrt{3} \\ & =5-\sqrt{2}-\sqrt{3} \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} \lim _{n \rightarrow \infty} & \frac{1}{n^2}\left[e^{1 / n}+2 e^{2 / n}+3 e^{3 / n}+\ldots+2 n e^2\right] \\ & =\lim _{n \rightarrow \infty} \frac{1}{n^2}\left[\sum_{k=1}^{2 n} k e^{k / n}\right] \\ & =\lim _{n \rightarrow \infty} \frac{1}{n}\left[\sum_{k=1}^{2 n}\left(\frac{k}{n}\right) e^{k / n}\right] \\ & =\int_0^2 x e^x d x=\left(x e^x-e^x\right)_0^2 \\ & =\left(2 e^2-e^2\right)-(0-1)=e^2+1 \end{aligned} \end{aligned} $

Given, $m, n, p, q$ be four positive inegers

$ \begin{aligned} & \int_0^{2 \pi} \sin ^m x \cos ^n x d x=4 \int_0^{\pi / 2} \sin ^m x \cos ^n x d x \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & \int_0^{2 \pi} \sin ^p x \cos ^n x d x=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ & \text { And } \int_0^{2 \pi} \sin ^p x \cos ^q x d x=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii) \\ & \quad a=m+n+p, b=m+n+q \end{aligned} $

From Eq. (i), we know that

$ \int_0^{2 \pi} f(x) d x=4 \int_0^{\pi / 2} f(x) $

If $f$ is periodic with $\pi$ and symmetric in all four quadrants

$\Rightarrow \quad m$ and $n$ are even

$ \int_0^{2 \pi} \sin ^p x \cos ^n x d x=0 $

$\Rightarrow$ Integrand is odd over symmetric interval $[0,2 \pi]$

Since, $n$ is even so, $p$ is odd also,

$ \begin{aligned} & \int_0^\pi \sin ^p x \cos ^q x d x=0 \\ & \Rightarrow \quad \sin ^p x \cos ^q x \text { is odd about } x=\frac{\pi}{2} \\ & \Rightarrow \quad q \text { is odd } \\ & \begin{aligned} \therefore & a=m+n+p \\ & =\text { even }+ \text { even }+ \text { odd }=\text { odd } \\ \text { And } b & =m+n+p \\ & =\text { even }+ \text { even }+ \text { odd }=\text { odd } \end{aligned} \end{aligned} $

$I=\int_0^2 \sqrt{(x+3)(2-x)} d x$

$ \begin{aligned} & \Rightarrow \quad \int_0^2 \sqrt{-x^2-x+6} d x \\ & \Rightarrow \quad \int_0^2 \sqrt{-\left(x^2+x-6\right)} d x \\ & \Rightarrow \quad \int_0^2 \sqrt{\frac{25}{4}-\left(x+\frac{1}{2}\right)^2} d x \end{aligned} $

Put $u=x+\frac{1}{2} \Rightarrow d u=d x$

When $x=0, u=\frac{1}{2}$ and $x=2$

$ \begin{aligned} & u=2+\frac{1}{2}=\frac{5}{2} \\ & I=\int_{\frac{1}{2}}^{\frac{5}{2}} \sqrt{\left(\frac{5}{2}\right)^2-u^2} d u \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad\left[\frac{u}{2} \sqrt{\frac{25}{4}-u^2}+\frac{\frac{25}{4}}{2} \sin ^{-1}\left(\frac{u}{\frac{5}{2}}\right)\right]_{\frac{1}{2}}^{\frac{5}{2}} \\ & \Rightarrow \quad\left[\frac{u}{2} \sqrt{\frac{25}{4}-u^2}+\frac{25}{8} \sin ^{-1}\left(\frac{2 u}{5}\right)\right]_{\frac{1}{2}}^{\frac{5}{2}} \\ & \Rightarrow \quad\left\{\frac{\frac{5}{2}}{2} \sqrt{\frac{25}{4}-\frac{25}{4}}+\frac{25}{8} \sin ^{-1}(1)\right\} -\left\{\frac{1}{4} \sqrt{\frac{25}{4}-\frac{1}{4}}+\frac{25}{8} \sin ^{-1}\left(\frac{1}{5}\right)\right\} \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad 0+\frac{25}{8} \cdot \frac{\pi}{2}-\frac{1}{4} \sqrt{6}-\frac{25}{8} \sin ^{-1}\left(\frac{1}{5}\right) \\ & \Rightarrow \quad \frac{25 \pi}{16}-\frac{\sqrt{6}}{4}-\frac{25}{8} \sin ^{-1}\left(\frac{1}{5}\right) \\ & \Rightarrow \quad \frac{25}{8}\left(\frac{\pi}{2}-\sin ^{-1}\left(\frac{1}{5}\right)\right)-\frac{\sqrt{6}}{4} \\ & \Rightarrow \quad \frac{25}{8} \cos ^{-1}\left(\frac{1}{5}\right)-\frac{\sqrt{6}}{4} \end{aligned} $

$\int_0^{\frac{\pi}{4}} x^2 \sin 2 x d x, I=\int x^2 \sin 2 x d x$

Put $t=2 x \Rightarrow d t=2 \cdot d x$

So, $I=\int \frac{t^2 \sin t}{8} d t=\frac{1}{8} \int t^2 \sin t d t$

$ \begin{aligned} & =\frac{1}{8}\left[t^2 \cdot(-\cos t)-1 \cdot(-2) \cdot \int t \cos t d t\right] \\ & =\frac{1}{8}\left[-t^2 \cos t+2 \int t \cdot \cos t d t\right] \\ & =\frac{1}{8}\left[-t^2 \cos t+2(t \cdot \sin t-(-\cos t))\right] \\ & =\frac{1}{8}\left[-t^2 \cos t+2 t \sin t+2 \cos t\right] \end{aligned} $

$ \begin{aligned} & =\frac{1}{8}\left[(2 x)^2 \cos (2 x)+2(2 x \cdot \sin (2 x))\right. +2 \cos (2 x)] \\ & =\frac{-x^2 \cos (2 x)+x \sin (2 x)}{2}+\frac{\cos (2 x)}{4} \end{aligned} $

$ \begin{aligned} & \text { So, } \int_0^{\frac{\pi}{4}} x^2 \sin 2 x d x \\ & =\left[\frac{-x^2 \cos (2 x)+x \sin (2 x)}{2}+\frac{\cos (2 x)}{4}\right]_0^{\frac{\pi}{4}} \\ & =\left\{\frac{-\left(\frac{\pi}{4}\right)^2 \cos \left(2 \times \frac{\pi}{4}\right)+\frac{\pi}{4} \sin \left(2 \times \frac{\pi}{4}\right)}{2}\right. \left.+\frac{\cos \left(\frac{2 \pi}{4}\right)}{4}\right\}-\left\{\frac{0}{2}+\frac{\cos (0)}{4}\right\} \\ & \Rightarrow\left(0+\frac{\pi}{8} \cdot 1+\frac{1}{4} \cdot 0\right)-\left(0+\frac{1}{4}\right) \\ & \quad=\frac{\pi}{8}-\frac{1}{4}=\frac{\pi-2}{8} \end{aligned} $

$ \begin{aligned} & \int_{-2 \pi}^{2 \pi} \sin ^4 x \cos ^6 x d x \\ & f(x)=\sin ^4 x \cos ^6 x \end{aligned} $

Since, $\sin x$ is an odd function and $\sin ^2 x$ is an even function and $\cos ^6 x$ is an even function.

So, $f(x)$ is an even function.

$ \begin{gathered} \therefore \int_{-2 \pi}^{2 \pi} \sin ^4 x \cos ^6 x d x=2 \int_0^{2 \pi} \sin ^4 x \cos ^6 x d x \\ {\left[\because \int_{-a}^a f(x) d x=2 \int_0^a f(x) d x\right]} \\ =2 \times 2 \int_0^\pi \sin ^4 x \cos ^6 x d x \\ {\left[\because \int_0^{n \pi} f(x) d x=n \int_0^T f(x) d x\right]} \\ =4 \int_0^\pi \sin ^4 x \cos ^6 x d x \\ =4 \times 2 \int_0^{\frac{\pi}{2}} \sin ^4 x \cos ^6 x d x \\ {\left[\because \int_0^{2 a} f(x) d x=2 \int_0^a f(x) d x\right]} \end{gathered} $

$ \begin{aligned} &=8 \int_0^{\frac{\pi}{2}} \sin ^4 x \cos ^6 x d x\\ &[\because \text { If } m \text { and } n \text { are both even, then }\\ &\int_0^{\frac{\pi}{2}} \sin ^m x \cos ^n x d x= \end{aligned} $

$ \left.\frac{[(m-1)(m-3) \ldots 2 \text { or } 1][(n-1)(n-3) \ldots 2 \text { or } 1]}{[(m+n)(m+n-2) \ldots 2 \text { or } 1]} \times \frac{\pi}{2}\right] $

$ \begin{aligned} & =8 \times \frac{3 \times 1 \times 5 \times 3 \times 1}{10 \times 8 \times 6 \times 4 \times 2} \times \frac{\pi}{2} \\ & =\frac{3 \pi}{64} \end{aligned} $

$ \begin{aligned} & I=\int_0^1 x \sin ^{-1} x \cdot d x \\ & \because \int u v \cdot d x=u \int v d x-\int\left(u^{\prime} \int u d x\right) d x \end{aligned} $

Here, $u=\sin ^{-1} x, v=x$

$ \begin{aligned} &\because I=\left[\sin ^{-1} x \int x \cdot d x-\int\left(\frac{1}{\sqrt{1-x^2}} \int x \cdot d x\right) d x\right]_0^1 \\ & =\left[\sin ^{-1} x \cdot \frac{x^2}{2}-\int \frac{x^2}{2 \sqrt{1-x^2}} d x\right]_0^1 \\ & =\left(\sin ^{-1}(1) \cdot \frac{1}{2}-\sin ^{-1}(0) \cdot 0\right)-\int_0^1 \frac{x^2}{2 \sqrt{1-x^2}} d x \\ & =\frac{\pi}{4}-\frac{1}{2} \int_0^1 \frac{x^2}{\sqrt{1-x^2}} d x \end{aligned} $

Let $x=\sin \theta \Rightarrow d x=\cos \theta \cdot d \theta$ at $x=1 \Rightarrow \theta=\pi / 2$ and $x=0 \Rightarrow \theta=0$

$ \begin{aligned} & =\frac{\pi}{4}-\frac{1}{2} \int_0^{\pi / 2} \frac{\sin ^2 \theta \cdot \cos \theta}{\cos \theta} d \theta \\ & =\frac{\pi}{4}-\frac{1}{2} \int_0^{\pi / 2} \sin ^2 \theta \cdot d \theta \\ & =\frac{\pi}{4}-\frac{1}{2} \int_0^{\pi / 2}\left(\frac{1}{2}-\frac{\cos 2 \theta}{2}\right) d \theta \\ & =\frac{\pi}{4}-\frac{1}{2}\left[\frac{\theta}{2}-\frac{\sin 2 \theta}{4}\right]_0^{\pi / 2} \\ & =\frac{\pi}{4}-\frac{1}{2}\left[\frac{\pi}{4}-0-0\right]=\frac{\pi}{4}-\frac{\pi}{8}=\frac{\pi}{8} \end{aligned} $

$\because I=\int_{-\pi / 2}^{\pi / 2} \sin (x-[x]) d x$

for $x \in[-\pi / 2,-1] \Rightarrow[x]=-2$

and $x-[x]=x+2$

for $x \in[-1,0) \Rightarrow[x]=-1$

and $x-[x]=x+1$

for $x \in[0,1] \Rightarrow[x]=0$ and $x-[x]=x$

for $x \in\left[1, \frac{\pi}{2}\right] \Rightarrow x-[x]=x-1$

$ \begin{aligned} \therefore I=\int_{-\pi / 2}^{-1} \sin (x & +2 d x+\int_{-1}^0 \sin (x+1) d x +\int_0^1 \sin x \cdot d x+\int_1^{\pi / 2} \sin (x-1) d x \end{aligned} $

$ \begin{aligned} = & {[-\cos (x+2)]_{-\pi / 2}^1+[-\cos (x+1)]_{-1}^0 } +[-\cos x]_0^1+[-\cos (x-1)]_1^{\pi / 2} \\ = & -\cos (1)+\cos (2-\pi / 2)-\cos (1)+\cos (0) -\cos (1)+\cos (0)-\cos (\pi / 2-1)+\cos (0) \\ = & -3 \cos (1)+3+\cos (-(\pi / 2-2)) -\cos (\pi / 2-1) \\ = & 3-3 \cos (1)+\sin (2)-\sin (1) \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { } \int_0^2 x^2(2-x)^5 d x \\ & =\int_0^2 x^2\left(32-80 x+80 x^2-40 x^3\right. \left.+10 x^4-x^5\right) d x \\ & =\left[\begin{array}{rl} \frac{32 x^3}{3}-\frac{80 x^4}{4}+\frac{80 x^5}{5}-\frac{40 x^6}{6} \left.+\frac{10 x^7}{7}-\frac{x^8}{8}\right]_0^2 \end{array}\right. \\ & =\left(\frac{32 \times 8}{3}-20 \times 16+16 \times 32-\frac{20}{3} \times 64+\frac{10}{7}\right. \left.\times 128-\frac{256}{8}\right)-0\\ & =\frac{256}{3}+16 \times 12-\frac{1280}{3}+\frac{1280}{7}-32 \\ & =\frac{-1024}{3}+\frac{1280}{7}+160 \\ & =\frac{-7168+3840+3360}{21}=\frac{32}{21} \end{aligned} \end{aligned} $

$\because f(x)=\max \left\{x^3-4, x^4-4\right\}$,

$ \begin{aligned} & g(x)=\min \left\{x^2, x^3\right\} \\ & \text { at } x=-1 \Rightarrow x^2=1, x^3=-1 \\ & \Rightarrow \min \{1,-1\}=-1=(-1)^3 \end{aligned} $

Similarly, at $x=0.5, \Rightarrow x^2=0.25$,

$ \begin{aligned} & x^3=0.125 \\ & \Rightarrow \min \{0.25,0.125\}=0.125=(0.5)^3 \end{aligned} $

So, for $x \in[-1,1] \Rightarrow g(x)=x^3$

Similarly, analysis of $f(x)$

When, $x^3-4=x^4-4$

$ \Rightarrow \quad x^3(x-1)=0 $

So, $x=0,1$

for $x<0 \Rightarrow x^4-4>x^3-4$

for $0x^4-x$

thus, $f(x)=\left\{\begin{array}{lc}x^4-4, & -1 \leq x \leq 0 \\ x^3-4, & 0 < x \leq 1\end{array}\right.$

$ \begin{aligned} & \text { Therefore, } f(x)-g(x) \\ & =\left\{\begin{array}{cc} \left(x^4-4\right)-x^3=x^4-x^3-4, & -1 \leq x \leq 0 \\ \left(x^3-4\right)-x^3=-4, & 0 < x \leq 1 \end{array}\right. \\ & \therefore \int_{-1}^1(f(x)-g(x)) d x=\int_{-1}^0\left(x^4-x^3-4\right) d x +\int_0^1-4 \cdot d x \end{aligned} $

$ \begin{aligned} & =\left[\frac{x^5}{5}-\frac{x^4}{4}-4 x\right]_{-1}^0-4[x]_0^1 \\ & =\left[0-\left(-\frac{1}{5}-\frac{1}{4}+4\right)-4(1)-0\right] \end{aligned} $

$ =-8+\frac{1}{5}+\frac{1}{4}=\frac{-160+5+4}{20}=\frac{-151}{20} $

$ \begin{aligned} & \text { Let } I=\int_0^1 \frac{2 x+5}{x^2+3 x+2} d x \\ & \begin{aligned} \frac{2 x+5}{x^2+3 x+2} & =\frac{A}{(x+1)}+\frac{B}{(x+2)} \\ 2 x+5 & =A(x+2)+B(x+1) \\ 2 & =A+B, 5=2 A+B \end{aligned} \end{aligned} $

Solving we get $A=3, B=-1$

$ \begin{aligned} I & =\int_0^1\left(\frac{3}{x+1}-\frac{1}{x+2}\right) d x \\ & =3[\ln \mid(x+1)]_0^1-[\ln |x+2|]_0^1 \\ & =3 \ln 2-\ln 3+\ln 2 \\ & =4 \ln 2-\ln 3 \\ & =4 \log 2-\log 3=\log \left(\frac{2^4}{3}\right) \\ & =\log \left(\frac{16}{3}\right) \end{aligned} $

Let $I=\int_0^1 x^{\frac{5}{2}}(1-x)^{\frac{3}{2}} d x$

$ \text { } \begin{aligned}Put\,\,\,\, & x=\sin ^2 \theta \\ & d x=2 \sin \theta \cos \theta d \theta \\ \Rightarrow I= & \int_0^{\frac{\pi}{2}} \sin ^5 \theta \cdot\left(\cos ^2 \theta\right)^{\frac{3}{2}} \cdot 2 \sin \theta \cdot \cos \theta d \theta \\ = & 2 \int_0^{\frac{\pi}{2}} \sin ^6 \theta \cdot \cos ^4 \theta d \theta \end{aligned} $

Using Wali's theorem,

$ =2\left(\frac{5 \cdot 3 \cdot 1 \cdot 3 \cdot 1}{10 \cdot 8 \cdot 6 \cdot 4 \cdot 2} \cdot \frac{\pi}{2}\right)=\frac{3 \pi}{256} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} \lim _{n \rightarrow-\infty} \frac{1}{n^2} \cdot \sec ^2 \frac{1}{n^2}+ & \frac{2}{n^2} \sec ^2 \frac{4}{n^2} +\ldots+\frac{1}{n^2} \sec ^2 1 \end{aligned} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & =\lim _{n \rightarrow-\infty} \sum_{r=1}^n \frac{r}{n^2} \cdot \sec ^2 \frac{r^2}{n^2} \\ & =\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^n\left(\frac{r}{n}\right) \sec ^2\left(\frac{r}{n}\right)^2 \\ & =\int_0^1 x \cdot \sec ^2 x^2 d x \\ & 1 \leq r \leq n \\ & \frac{1}{n} \leq \frac{r}{n} \leq 1 \\ \because \quad & n \rightarrow \infty \\ & 0 \leq \frac{r}{n}<1 \end{aligned}\\ &\text { Put } \quad x^2=t\\ &\begin{aligned} & \Rightarrow \quad 2 x d x=d t \\ & \int_0^1 \sec ^2 t \frac{d t}{2}=\frac{1}{2}[\tan t]_0^1 \\ & \Rightarrow \quad \frac{\tan 1}{2} \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { Let } I=\int_0^\pi\left(\sin ^5 x \cos ^3 x+\sin ^4 x\right. \\ & \begin{array}{r} \left.\cos ^4 x+\sin ^3 x \cos ^4 x\right) d x \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ \Rightarrow I=\int_0^\pi\left[\sin ^5(\pi-x) \cos ^3(\pi-x)\right. \\ +\sin ^4(\pi-x) \cos ^4(\pi-x) \\ \left.+\sin ^3(\pi-x) \cdot \cos ^4(\pi-x)\right] d x \\ \Rightarrow I=\int_0^\pi-\sin ^5 x \cos ^3 x+\sin ^4 x \cos ^4 x \\ +\sin ^3 x \cos ^4 x d x \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{array} \end{aligned} $

On adding Eqs. (i) and (ii), we get

$ \begin{aligned} & 2 I=2 \int_0^\pi\left(\sin ^4 x \cos ^4 x+\sin ^3 x \cos ^4 x\right) d x \\ & I=\int_0^\pi \frac{\sin ^4 2 x}{16} d x+\int_0^\pi \sin ^3 x \cdot \cos ^4 x d x \end{aligned} $

$ \begin{aligned} & \cos x=t \\ & \Rightarrow-\sin x d x=d t \\ & =\frac{1}{16} \int_0^\pi \frac{(1-\cos 4 x)^2}{4} d x+\int_1^{-1} t^4 \cdot\left(1-t^2\right)(-d t) \\ & =\frac{1}{64} \int_0^\pi\left(1-2 \cos 4 x+\cos ^2 4 x\right) d x \text {. } +\int_1^{-1} t^4\left(t^2-1\right) d t \\ & =\frac{1}{64} \int_0^\pi\left(1-2 \cos 4 x+\frac{1+\cos 8 x}{2}\right) d x +\left(\frac{t^7}{5}-\frac{t^5}{5}\right)_1^{-1} \\ & =\frac{1}{64} \int_0^\pi\left(\frac{3}{2}-2 \cos 4 x+\frac{1}{2} \cos 8 x\right) d x +\left(\frac{2}{5}-\frac{2}{7}\right) \\ & =\frac{3}{128}[x]_0^\pi-\frac{2}{64}\left[\frac{\sin 4 x}{4}\right]_0^\pi+\frac{1}{128}\left[\frac{\sin 8 x}{8}\right]_0^\pi +\frac{4}{35} \\ & \Rightarrow \frac{3 \pi}{128}+\frac{4}{35} \end{aligned} $