Definite Integration

To solve the given integral $\int\limits_0^{\frac{\pi}{3}} \cos ^4 x \mathrm{~d} x$, we'll apply a known power-reduction formula that allows us to express even powers of sine and cosine functions in terms of cosine of multiple angles. Specifically for $\cos^4 x$, we can write it in terms of double angles as:

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

We can then expand and simplify the integral using this formula. Let's proceed with this:

$ \int\limits_0^{\frac{\pi}{3}} \cos^4 x \mathrm{~d}x = \int\limits_0^{\frac{\pi}{3}} \left(\frac{1 + \cos(2x)}{2}\right)^2 \mathrm{~d}x $

Now, let's expand the integrand and then integrate term by term:

$ = \int\limits_0^{\frac{\pi}{3}} \left(\frac{1}{4} + \frac{1}{2} \cos(2x) + \frac{1}{4} \cos^2(2x)\right)\mathrm{~d}x $

For the $\cos^2(2x)$ term, we again use the power reduction formula:

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

Let's substitute this into the integral and continue:

$ = \int\limits_0^{\frac{\pi}{3}} \left(\frac{1}{4} + \frac{1}{2} \cos(2x) + \frac{1}{4} \left(\frac{1 + \cos(4x)}{2}\right)\right)\mathrm{~d}x $

Simplify and integrate:

$ = \int\limits_0^{\frac{\pi}{3}} \left(\frac{1}{4} + \frac{1}{2} \cos(2x) + \frac{1}{8} + \frac{1}{8} \cos(4x)\right)\mathrm{~d}x $

$ = \int\limits_0^{\frac{\pi}{3}} \left(\frac{3}{8} + \frac{1}{2} \cos(2x) + \frac{1}{8} \cos(4x)\right)\mathrm{~d}x $

$ = \left. \left(\frac{3}{8} x + \frac{1}{4} \sin(2x) + \frac{1}{32} \sin(4x)\right) \right|_0^{\frac{\pi}{3}} $

Evaluating this from $0$ to $\frac{\pi}{3}$:

$ = \left(\frac{3}{8} \cdot \frac{\pi}{3} + \frac{1}{4} \sin\left(2 \cdot \frac{\pi}{3}\right) + \frac{1}{32} \sin\left(4 \cdot \frac{\pi}{3}\right)\right) - \left(\frac{3}{8} \cdot 0 + \frac{1}{4} \sin(0) + \frac{1}{32} \sin(0)\right) $

$ = \frac{\pi}{8} + \frac{1}{4} \sin\left(\frac{2\pi}{3}\right) + \frac{1}{32} \sin\left(\frac{4\pi}{3}\right) $

$\sin\left(\frac{2\pi}{3}\right)$ is $\frac{\sqrt{3}}{2}$ and $\sin\left(\frac{4\pi}{3}\right)$ is $-\frac{\sqrt{3}}{2}$:

$ = \frac{\pi}{8} + \frac{1}{4} \cdot \frac{\sqrt{3}}{2} - \frac{1}{32} \cdot \frac{\sqrt{3}}{2} $

$ = \frac{\pi}{8} + \frac{\sqrt{3}}{8} - \frac{\sqrt{3}}{64} $

Now, combining terms we get the final result:

$ = \frac{\pi}{8} + \frac{8\sqrt{3} - \sqrt{3}}{64} $

$ = \frac{\pi}{8} + \frac{7\sqrt{3}}{64} $

Now, let's match this result to the form $\mathrm{a}\pi + \mathrm{b}\sqrt{3}$ and find $a$ and $b$:

$ a = \frac{1}{8}, \quad b = \frac{7}{64} $

Now we find $9a + 8b$:

$ 9a + 8b = 9 \cdot \frac{1}{8} + 8 \cdot \frac{7}{64} $

$ = \frac{9}{8} + \frac{7}{8} $

$ = \frac{16}{8} $

$ = 2 $

Therefore, the value of $9a + 8b$ is 2, which correspond to Option A.

$\begin{aligned} & I=\int_0^1\left(2 x^3-3 x^2-x+1\right)^{1 / 3} d x \\\\ & I=\int_0^1\left((2 x-1)\left(x^2-x-1\right)\right)^{1 / 3} d x \\\\ & I=\int_0^1\left[(2(1-x)-1)\left((1-x)^2-(1-x)-1\right)\right]^{1 / 3} d x \\\\ & I=\int_0^1\left((1-2 x)\left(x^2-x-1\right)\right)^{1 / 3} d x\end{aligned}$

$\begin{aligned} & I=-\int_0^1\left((2 x-1)\left(x^2-x-1\right)\right)^{1 / 3} d x \\\\ & I=-I \\\\ & 2 I=0 \\\\ & I=0\end{aligned}$

Take $I=\int\limits_0^{\pi / 4} \frac{x d x}{\sin ^4(2 x)+\cos ^4(2 x)}$

Let $2 x=t$

$2 d x=d t$

$d x=\frac{d t}{2}$

$\begin{aligned} & I=\int\limits_0^{\pi / 2} \frac{t / 2 \cdot 1 / 2 d t}{\sin ^4 t+\cos ^4 t} \\\\ & I=\frac{1}{4} \int\limits_0^{\pi / 2} \frac{t d t}{\sin ^4 t+\cos ^4 t} \\\\ & =\frac{1}{4} \int\limits_0^{\pi / 2} \frac{\left(\frac{\pi}{2}-t\right) d t}{\sin ^4(\pi / 2-t)+\cos ^4(\pi / 2-t)} \\\\ & =\frac{1}{4} \int\limits_0^{\pi / 2} \frac{\left(\frac{\pi}{2}-t\right)}{\sin ^4 t+\cos ^4 t}\end{aligned}$

$\begin{aligned} & 2 I=\frac{1}{4} \int\limits_0^{\pi / 2} \frac{\frac{\pi}{2}}{\sin ^4 t+\cos ^4 t} d t \\\\ & 2 I=\frac{\pi}{8} \int\limits_0^{\pi / 2} \frac{d t}{\sin ^4 t+\cos ^4 t} \\\\ & 2 I=\frac{\pi}{8} \int\limits_0^{\pi / 2} \frac{\sec ^4 t}{1+\tan ^4 t} d t\end{aligned}$

$\begin{aligned} & \text { Put tant }=y \\\\ & \sec ^2 t d t=d y \\\\ & 2 I=\frac{\pi}{8} \int\limits_0^{\infty} \frac{\left(1+y^2\right) d y}{1+y^4} \\\\ & =\frac{\pi}{8} \int\limits_0^{\infty} \frac{1+\frac{1}{y^2}}{y^2+\frac{1}{y^2}-2+2} d y \\\\ & = \frac{\pi}{8} \int\limits_0^{\infty} \frac{\left(1+\frac{1}{y^2}\right) d y}{2+\left(y-\frac{1}{y}\right)^2}\end{aligned}$

$\begin{aligned} & \text { Put, } y-\frac{1}{y}=4 \\\\ & 2 I=\frac{\pi}{8} \int\limits_{-\infty}^{\infty} \frac{d u}{2+u^2} \\\\ & =\frac{\pi}{8 \sqrt{2}}\left[\tan ^{-1} \frac{y}{\sqrt{2}}\right]_{-\infty}^{\infty} \\\\ & =\frac{\sqrt{2} \pi^2}{32}\end{aligned}$

$\begin{aligned} & \mathrm{f}(\mathrm{x})=\int_\limits{-\mathrm{x}}^{\mathrm{x}}\left(|\mathrm{t}|-\mathrm{t}^2\right) \mathrm{e}^{-\mathrm{t}^2} \mathrm{dt} \\ & \Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=2 \cdot\left(|\mathrm{x}|-\mathrm{x}^2\right) \mathrm{e}^{-\mathrm{x}^2} \ldots \ldots \ldots . .(1) \\ & \mathrm{g}(\mathrm{x})=\int_\limits0^{\mathrm{x}^2} \mathrm{t}^{\frac{1}{2}} \mathrm{e}^{-t} \mathrm{dt} \\ & \mathrm{g}^{\prime}(\mathrm{x})=\mathrm{xe}^{-\mathrm{x}^2}(2 \mathrm{x})-0 \\ & \mathrm{f}^{\prime}(\mathrm{x})+\mathrm{g}^{\prime}(\mathrm{x})=2 \mathrm{xe}^{-\mathrm{x}^2}-2 \mathrm{x}^2 \mathrm{e}^{-\mathrm{x}^2}+2 \mathrm{x}^2 \mathrm{e}^{-\mathrm{x}^2} \end{aligned}$

Integrating both sides w.r.t. $\mathrm{x}$

$\begin{aligned} & \mathrm{f}(\mathrm{x})+\mathrm{g}(\mathrm{x})=\int_\limits0^\alpha 2 \mathrm{xe}^{-\mathrm{x}^2} \mathrm{dx} \\ & \mathrm{x}^2=\mathrm{t} \\ & \Rightarrow \int_0^{\sqrt{\alpha}} \mathrm{e}^{-\mathrm{t}} \mathrm{dt}=\left[-\mathrm{e}^{-\mathrm{t}}\right]_0^{\sqrt{\alpha}} \\ & =-\mathrm{e}^{\left(\log _c(9)^{-1}\right)+1} \\ & \Rightarrow 9(\mathrm{f}(\mathrm{x})+\mathrm{g}(\mathrm{x}))=\left(1-\frac{1}{9}\right) 9=8 \end{aligned}$

$\begin{aligned} &f(x)=\frac{x}{\left(1+x^4\right)^{1 / 4}}\\ &f \circ f(x)=\frac{f(x)}{\left(1+f(x)^4\right)^{1 / 4}}=\frac{\frac{x}{\left(1+x^4\right)^{1 / 4}}}{\left(1+\frac{x^4}{1+x^4}\right)^{1 / 4}}=\frac{x}{\left(1+2 x^4\right)^{1 / 4}}\\ &f(f(f(f(x))))=\frac{x}{\left(1+4 x^4\right)^{1 / 4}} \end{aligned}$

$18 \int_\limits0^{\sqrt{2 \sqrt{5}}} \frac{x^3}{\left(1+4 x^4\right)^{1 / 4}} d x$

$\begin{aligned} & \text { Let } 1+4 \mathrm{x}^4=\mathrm{t}^4 \\ & 16 \mathrm{x}^3 \mathrm{dx}=4 \mathrm{t}^3 \mathrm{dt} \\ & \frac{18}{4} \int_\limits1^3 \frac{\mathrm{t}^3 \mathrm{dt}}{\mathrm{t}} \\ & =\frac{9}{2}\left(\frac{\mathrm{t}^3}{3}\right)_1^3 \\ & =\frac{3}{2}[26]=39 \end{aligned}$

$\begin{aligned} & y=f(x) \Rightarrow \frac{d y}{d x}=f^{\prime}(x) \\ & \left.\frac{d y}{d x}\right)_{(1, f(1))}=f^{\prime}(1)=\tan \frac{\pi}{6}=\frac{1}{\sqrt{3}} \Rightarrow f^{\prime}(1)=\frac{1}{\sqrt{3}} \\ & \left.\frac{d y}{d x}\right)_{(3, f(3))}=f^{\prime}(3)=\tan \frac{\pi}{4}=1 \Rightarrow f^{\prime}(3)=1 \\ & 27 \int_\limits1^3\left(\left(f^{\prime}(t)\right)^2+1\right) f^{\prime \prime}(t) d t=\alpha+\beta \sqrt{3} \\ & I=\int_\limits1^3\left(\left(f^{\prime}(t)\right)^2+1\right) f^{\prime \prime}(t) d t \\ & f^{\prime}(t)=z \Rightarrow f^{\prime \prime}(t) d t=d z \\ & z=f^{\prime}(3)=1 \\ & z=f^{\prime}(1)=\frac{1}{\sqrt{3}} \end{aligned}$

$\begin{aligned} & I=\int_\limits{1 / \sqrt{3}}^1\left(z^2+1\right) d z=\left(\frac{z^3}{3}+z\right)_{1 / \sqrt{3}}^1 \\ & =\left(\frac{1}{3}+1\right)-\left(\frac{1}{3} \cdot \frac{1}{3 \sqrt{3}}+\frac{1}{\sqrt{3}}\right) \\ & =\frac{4}{3}-\frac{10}{9 \sqrt{3}}=\frac{4}{3}-\frac{10}{27} \sqrt{3} \\ & \alpha+\beta \sqrt{3}=27\left(\frac{4}{3}-\frac{10}{27} \sqrt{3}\right)=36-10 \sqrt{3} \\ & \alpha=36, \beta=-10 \\ & \alpha+\beta=36-10=26 \end{aligned}$

To determine the integral of the piecewise function $f$ over the interval $[-2, 2]$, we first ensure that $f$ is differentiable on $\mathbb{R}$, as given in the problem statement. Differentiability implies continuity, so $f$ must also be continuous at $x=1$.

The condition for continuity at $x=1$ is:

$x^2 + 3x + a = bx + 2$ at $x=1$.

This simplifies to:

$1 + 3 + a = b(1) + 2$

$\Rightarrow a = b - 2$

The first derivative of $f$ gives us two different expressions depending on the value of $x$:

For $x \leq 1$, $f'(x) = 2x + 3$; and for $x > 1$, $f'(x) = b$.

For $f$ to be differentiable at $x=1$, these derivatives must be equal at that point. Setting $f'(1)$ from both expressions equal to each other gives $2(1) + 3 = b$, thus $b = 5$. And from $a = b - 2$, we have $a = 3$.

Now, we can calculate the integral of $f$ over the specified interval:

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

$\begin{aligned} & =\left[\frac{x^3}{3}+\frac{3 x^2}{2}+3 x\right]_{-2}^1+\left[\frac{5 x^2}{2}+2 x\right]_1^2 \\\\ & =\left(\frac{1}{3}+\frac{3}{2}+3\right)-\left(\frac{-8}{3}+6-6\right)+\left(10+4-\frac{5}{2}-2\right) \\\\ & =6+\frac{3}{2}+12-\frac{5}{2}=17 \end{aligned}$

$\begin{array}{ll} \mathrm{f}(\mathrm{x})=a \mathrm{e}^{2 \mathrm{x}}+b \mathrm{e}^{\mathrm{x}}+\mathrm{cx} & \mathrm{f}(0)=-1 \\\\ & \mathrm{a}+\mathrm{b}=-1 \\\\ \mathrm{f}^{\prime}(\mathrm{x})=2 a \mathrm{e}^{2 \mathrm{x}}+b \mathrm{e}^{\mathrm{x}}+\mathrm{c} & \mathrm{f}^{\prime}(\ln 2)=21 \\\\ & 8 \mathrm{a}+2 \mathrm{~b}+\mathrm{c}=21 \\\\ \int_\limits0^{\ln 4}\left(a \mathrm{e}^{2 \mathrm{x}}+b \mathrm{e}^{\mathrm{x}}\right) \mathrm{dx}=\frac{39}{2} & \end{array}$

$\begin{aligned} & {\left[\frac{\mathrm{ae}^{2 \mathrm{x}}}{2}+\mathrm{be}^{\mathrm{x}}\right]_0^{\ln 4}=\frac{39}{2} \Rightarrow 8 \mathrm{a}+4 \mathrm{~b}-\frac{\mathrm{a}}{2}-\mathrm{b}=\frac{39}{2}} \\\\ & 15 a+6 b=39 \\\\ & 15 a-6 a-6=39 \\\\ & 9 \mathrm{a}=45 \Rightarrow \mathrm{a}=5 \\\\ & b=-6 \\\\ & c=21-40+12=-7 \\\\ & \mathrm{a}+\mathrm{b}+\mathrm{c}-8 \\\\ & |\mathrm{a}+\mathrm{b}+\mathrm{c}|=8 \\\\ \end{aligned}$

$\begin{aligned} & \lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{n^3}{n^4\left(1+\frac{k^2}{n^2}\right)\left(1+\frac{3 k^2}{n^2}\right)} \\ & =\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^n \frac{n^3}{\left(1+\frac{k^2}{n^2}\right)\left(1+\frac{3 k^2}{n^2}\right)} \end{aligned}$

$=\int_\limits0^1 \frac{d x}{3\left(1+x^2\right)\left(\frac{1}{3}+x^2\right)}$

$\begin{aligned} & =\int_\limits0^1 \frac{1}{3} \times \frac{3}{2} \frac{\left(x^2+1\right)-\left(x^2+\frac{1}{3}\right)}{\left(1+x^2\right)\left(x^2+\frac{1}{3}\right)} d x \\ & =\frac{1}{2} \int_\limits0^1\left[\frac{1}{x^2+\left(\frac{1}{\sqrt{3}}\right)^2}-\frac{1}{1+x^2}\right] d x \\ & =\frac{1}{2}\left[\sqrt{3} \tan ^{-1}(\sqrt{3} x)\right]_0^1-\frac{1}{2}\left(\tan ^{-1} x\right)_0^1 \\ & =\frac{\sqrt{3}}{2}\left(\frac{\pi}{3}\right)-\frac{1}{2}\left(\frac{\pi}{4}\right)=\frac{\pi}{2 \sqrt{3}}-\frac{\pi}{8} \\ & =\frac{13 \pi}{8 \cdot(4 \sqrt{3}+3)} \end{aligned}$

$\begin{aligned} & \lim _{x \rightarrow 0} \frac{x \int_0^x f(t) d t}{\left(\frac{e^{x^2}-1}{x^2}\right) \times x^2} \\\\ & \lim _{x \rightarrow 0} \frac{\int_0^x f(t) d t}{x} \quad\left(\lim _{x \rightarrow 0} \frac{e^{x^2}-1}{x^2}=1\right) \\\\ & =\lim _{x \rightarrow 0} \frac{f(x)}{1} \text { (using L Hospital) } \\\\ & f(0)=\frac{1}{2} \\\\ & \alpha=\frac{1}{2} \\\\ & 8 \alpha^2=2 \end{aligned}$

Using L'hospital rule

$\begin{aligned} & =\lim _\limits{x \rightarrow \frac{\pi^{-}}{2}} \frac{0-\cos x \times 3 x^2}{2\left(x-\frac{\pi}{2}\right)} \\ & =\lim _\limits{x \rightarrow \frac{\pi^{-}}{2}} \frac{\sin \left(x-\frac{\pi}{2}\right)}{2\left(x-\frac{\pi}{2}\right)} \times \frac{3 \pi^2}{4} \\ & =\frac{3 \pi^2}{8} \end{aligned}$

$\begin{aligned} & I=\int_\limits{-\pi / 2}^{\pi / 2}\left(\frac{x^2 \cos x}{1+\pi^x}+\frac{1+\sin ^2 x}{1+e^{\sin x^{2023}}}\right) d x \\ & I=\int_\limits{-\pi / 2}^{\pi / 2}\left(\frac{x^2 \cos x}{1+\pi^{-x}}+\frac{1+\sin ^2 x}{1+e^{\sin (-x)^{2023}}}\right) d x \end{aligned}$

On Adding, we get

$2 I=\int_\limits{-\pi / 2}^{\pi / 2}\left(x^2 \cos x+1+\sin ^2 x\right) d x$

On solving

$\begin{aligned} & \mathrm{I}=\frac{\pi^2}{4}+\frac{3 \pi}{4}-2 \\ & \mathrm{a}=3 \end{aligned}$

$\begin{aligned} & I=\int_\limits0^\pi \frac{d x}{1-2 a \cos x+a^2} ; 0< a<1 \\ & I=\int_\limits0^\pi \frac{d x}{1+2 a \cos x+a^2} \\ & 2 I=2 \int_\limits0^{\pi / 2} \frac{2\left(1+a^2\right)}{\left(1+a^2\right)^2-4 a^2 \cos ^2 x} d x \\ & \Rightarrow I=\int_\limits0^{\pi / 2} \frac{2\left(1+a^2\right) \cdot \sec ^2 x}{\left(1+a^2\right)^2 \cdot \sec ^2 x-4 a^2} d x \\ & \Rightarrow I=\int_\limits0^{\pi / 2} \frac{2 \cdot\left(1+a^2\right) \cdot \sec ^2 x}{\left(1+a^2\right)^2 \cdot \tan ^2 x+\left(1-a^2\right)^2} d x \end{aligned}$

$\begin{aligned} & \Rightarrow \mathrm{I}=\int_\limits0^{\pi / 2} \frac{\frac{2 \cdot \sec ^2 \mathrm{x}}{1+\mathrm{a}^2} \cdot \mathrm{dx}}{\tan ^2 \mathrm{x}+\left(\frac{1-\mathrm{a}^2}{1+\mathrm{a}^2}\right)^2} \\ & \Rightarrow \mathrm{I}=\frac{2}{\left(1-\mathrm{a}^2\right)}\left[\frac{\pi}{2}-0\right] \\ & \mathrm{I}=\frac{\pi}{1-\mathrm{a}^2} \end{aligned}$

$\begin{aligned} & \int_\limits0^1 \frac{1}{\sqrt{3+x}+\sqrt{1+x}} d x=\int_\limits0^1 \frac{\sqrt{3+x}-\sqrt{1+x}}{(3+x)-(1+x)} d x \\ & \frac{1}{2}\left[\int_\limits0^1 \sqrt{3+x} d x-\int_\limits0^1(\sqrt{1+x}) d x\right] \end{aligned}$

$\begin{aligned} & \frac{1}{2}\left[2 \frac{(3+x)^{\frac{3}{2}}}{3}-\frac{2(1+x)^{\frac{3}{2}}}{3}\right]_0^1 \\ & \frac{1}{2}\left[\frac{2}{3}(8-3 \sqrt{3})-\frac{2}{3}\left(2^{\frac{3}{2}}-1\right)\right] \\ & \frac{1}{3}[8-3 \sqrt{3}-2 \sqrt{2}+1] \\ & =3-\sqrt{3}-\frac{2}{3} \sqrt{2}=a+b \sqrt{2}+c \sqrt{3} \\ & a=3, b=-\frac{2}{3}, c=-1 \\ & 2 a+3 b-4 c=6-2+4=8 \end{aligned}$

Equation of CE

$\begin{aligned} & y-1=-(x-3) \\ & x+y=4 \end{aligned}$

orthocentre lies on the line $x+y=4$

so, $a+b=4$

$I_1=\int_\limits a^b x \sin (x(4-x)) d x\quad$ ..... (i)

Using king rule

$I_1=\int_\limits a^b(4-x) \sin (x(4-x)) d x\quad$ .... (ii)

$\begin{aligned} & \text { (i) }+ \text { (ii) } \\ & 2 \mathrm{I}_1=\int_\limits{\mathrm{a}}^{\mathrm{b}} 4 \sin (\mathrm{x}(4-\mathrm{x})) \mathrm{dx} \\ & 2 \mathrm{I}_1=4 \mathrm{I}_2 \\ & \mathrm{I}_1=2 \mathrm{I}_2 \\ & \frac{\mathrm{I}_1}{\mathrm{I}_2}=2 \\ & \frac{36 \mathrm{I}_1}{\mathrm{I}_2}=72 \end{aligned}$

1. The given integral is:

$ I=\int_0^1 \frac{dx}{\left(5+2x-2x^2\right)\left(1+e^{2-4x}\right)} $ .............(i)

1. Perform a substitution, $x \rightarrow 1-x$. This gives:

$ I=\int_0^1 \frac{dx}{\left(5+2(1-x)-2(1-x)^2\right)\left(1+e^{2-4(1-x)}\right)} $

1. Simplify the expression inside the integral:

$ I=\int_0^1 \frac{dx}{\left(5+2-2x-2(1-2x+x^2)\right)\left(1+e^{4x -2}\right)} $ ............(ii)

1. Add the original integral (i) and the integral after substitution (ii):

$ 2I=\int_0^1 \frac{dx}{5+2x-2x^2} $

1. Factor the quadratic expression in the denominator:

$ 2I=\int_0^1 \frac{dx}{2\left(\frac{11}{4}-\left(x-\frac{1}{2}\right)^2\right)} $

1. To solve this integral, we can perform a change of variables using the substitution
   
   $x-\frac{1}{2}= \frac{\sqrt{11}}{2}\tan{u}$, then $dx=\frac{\sqrt{11}}{2}\sec^2{u} du$:

$ I=\frac{1}{\sqrt{11}}\int \frac{\sec^2{u}}{1+\tan^2{u}}du $

1. Using the identity $\sec^2{u}=1+\tan^2{u}$:

$ I=\frac{1}{\sqrt{11}}\int du = \frac{1}{\sqrt{11}}(u+C) $

1. Now we need to find the limits of the integral after the substitution. If $x=0$, then $u=\tan^{-1}\left(-\frac{1}{\sqrt{11}}\right)$. If $x=1$, then $u=\tan^{-1}\left(\frac{1}{\sqrt{11}}\right)$. So, the integral becomes:

$ I=\frac{1}{\sqrt{11}}\left[\tan^{-1}\left(\frac{1}{\sqrt{11}}\right)-\tan^{-1}\left(-\frac{1}{\sqrt{11}}\right)\right] $

1. Using the properties of the arctangent function, we can rewrite the integral as:

$ I=\frac{1}{\sqrt{11}} \ln\left(\frac{\sqrt{11}+1}{\sqrt{10}}\right) $

1. From this result, we have $\alpha=\sqrt{11}$ and $\beta=\sqrt{10}$. Now, we can find $\alpha^4 - \beta^4$:

$ \alpha^4 - \beta^4 = (11)^2 - (10)^2 = 121 - 100 = 21 $

Thus, $\alpha^4 - \beta^4 = 21$.

We're given the expression:

$\frac{e^{-\pi/4} + \int_0^{\pi / 4} e^{-x} \tan ^{50} x dx}{\int_0^{\pi / 4} e^{-x}(\tan x)^{49} dx + \int_0^{\pi / 4} e^{-x}(\tan x)^{51} dx}$

Notice that the integrals in the numerator and denominator have the same form. They both involve an integral of $e^{-x} \tan^n x$ from 0 to $\pi / 4$, where $n$ is an integer. Let's denote this integral as $I(n)$:

$I(n) = \int_0^{\pi / 4} e^{-x} \tan^n x dx$

We can then rewrite the original expression in terms of $I(n)$:

$\frac{e^{-\pi/4} + I(50)}{I(49) + I(51)}$

Now, we'll apply the method of integration by parts, which states that for two functions $u(x)$ and $v(x)$:

$\int u dv = uv - \int v du$

We'll choose:

$u = \tan^n x, \quad dv = e^{-x} dx$

Then we get:

$du = n \tan^{n-1} x \sec^2 x dx, \quad v = -e^{-x}$

Applying integration by parts, we have:

$I(n) = -e^{-x} \tan^n x \Bigg|_0^{\pi / 4} + n \int_0^{\pi / 4} e^{-x} \tan^{n-1} x \sec^2 x dx$

Since $\tan(\pi / 4) = 1$, the first term evaluates to:

$-e^{-\pi / 4}$

The second term becomes:

$n \int_0^{\pi / 4} e^{-x} \tan^{n-1} x (1 + \tan^2 x) dx = n \int_0^{\pi / 4} e^{-x} \tan^{n-1} x dx + n \int_0^{\pi / 4} e^{-x} \tan^{n+1} x dx$

This is equal to:

$n(I(n-1) + I(n+1))$

So we have:

$I(n) = -e^{-\pi / 4} + n(I(n-1) + I(n+1))$

Now we can substitute $n = 50$ into this equation:

$I(50) = -e^{-\pi / 4} + 50(I(49) + I(51))$

So the original expression becomes:

$\frac{e^{-\pi/4} + I(50)}{I(49) + I(51)} = \frac{e^{-\pi/4} - e^{-\pi / 4} + 50(I(49) + I(51))}{I(49) + I(51)} = 50$

$ \begin{aligned} & S_1: \lim _{n \rightarrow \infty} \frac{1}{n^2}[2+4+6+\ldots+2 n] \\\\ & \lim _{n \rightarrow \infty} 2 \frac{n(n+1)}{2 n^2}=1 \\\\ & S_2: \lim _{n \rightarrow \infty} \frac{1}{n^{16}}\left(\sum r^{15}\right)=\lim _{n \rightarrow \infty} \frac{1}{n} \sum\left(\frac{r}{n}\right)^{15} \\\\ & =\int_0^1 x^{15} d x=\frac{1}{16} \\\\ & \therefore \text { Both } S_1 \text { and } S_2 \text { are correct. } \end{aligned} $

$ \begin{aligned} & \mathrm{l}=\int_0^{\infty} \frac{6}{\left(\mathrm{e}^{\mathrm{x}}+1\right)\left(\mathrm{e}^{\mathrm{x}}+2\right)\left(\mathrm{e}^{\mathrm{x}}+3\right)} \mathrm{dx} \\\\ & =6 \int_0^{\infty}\left(\frac{\frac{1}{2}}{\mathrm{e}^{\mathrm{x}}+1}+\frac{-1}{\mathrm{e}^{\mathrm{x}}+2}+\frac{\frac{1}{2}}{\mathrm{e}^{\mathrm{x}}+3}\right) \mathrm{dx} \\\\ & =3 \int_0^{\infty} \frac{\mathrm{e}^{-\mathrm{x}}}{1+\mathrm{e}^{-\mathrm{x}}} \mathrm{dx}-6 \int_0^{\infty} \frac{\mathrm{e}^{-\mathrm{x}} \mathrm{dx}}{1+2 \mathrm{e}^{-\mathrm{x}}}+3 \int_0^{\infty} \frac{\mathrm{e}^{-\mathrm{x}}}{1+3 \mathrm{e}^{-\mathrm{x}}} \mathrm{dx} \\\\ & =3\left[-\ln \left(1+\mathrm{e}^{-\mathrm{x}}\right)\right]_0^{\infty}+6 \frac{1}{2}\left[\ln \left(1+2 \mathrm{e}^{-\mathrm{x}}\right)\right]_0^{\infty} -\frac{3}{3}\left[\ln \left(1+3 \mathrm{e}^{-\mathrm{x}}\right)\right]_0^{\infty} \\\\ & =3 \ln 2-3 \ln 3+\ln 4 \\\\ & =3 \ln \frac{2}{3}+\ln 4 \\\\ & =\ln \frac{32}{27} \end{aligned} $

The integral equation is given by :

$\int\limits_0^{\frac{\pi}{2}} f(\sin 2x) \sin x \, dx + \alpha \int\limits_0^{\frac{\pi}{4}} f(\cos 2x) \cos x \, dx = 0$

Step 1 : Break the first integral into two parts :

$I = \int\limits_0^{\frac{\pi}{4}} f(\sin 2x) \sin x \, dx + \int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} f(\sin 2x) \sin x \, dx + \alpha \int\limits_0^{\frac{\pi}{4}} f(\cos 2x) \cos x \, dx$

3. Apply the King's property, $\int\limits_a^b f(x) dx = \int\limits_a^b f(a+b-x) dx$, to the first integral and substitute $x - \frac{\pi}{4} = t$ in the second integral.

This gives :

$ \int\limits_0^{\frac{\pi}{4}} f(\cos 2x) \sin(\frac{\pi}{4}-x) \, dx + \int\limits_0^{\frac{\pi}{4}} f(\cos 2t) \sin(\frac{\pi}{4}+t) \, dt + \alpha \int\limits_0^{\frac{\pi}{4}} f(\cos 2x) \cos x \, dx = 0 $

$ \int\limits_0^{\frac{\pi}{4}} f(\cos 2x) [2 \sin \frac{\pi}{4} \cos x + \alpha \cos x] \, dx = 0 $

Then, noticing that $2 \sin \frac{\pi}{4} = \sqrt{2}$, you can factor out the term $\cos x$:

$ = (\alpha + \sqrt{2}) \int\limits_0^{\frac{\pi}{4}} f(\cos 2x) \cos x \, dx = 0 $

In order for this equation to hold true, either the integral of the function is zero, or the term outside the integral is zero. Since we have no reason to assume that the integral of the function is zero, we set the term outside the integral to zero, yielding the solution:

$ \alpha + \sqrt{2} = 0 \Rightarrow \alpha = -\sqrt{2} $

So, the correct answer to the original problem is $\alpha = -\sqrt{2}$, which corresponds to Option C.

$ \begin{aligned} \operatorname{Minimum}\left\{\mathrm{x}^2,\{\mathrm{x}\}\right\} & =\mathrm{x}^2 ; \mathrm{x} \in[0,1) \\\\ {\left[\mathrm{x}-\log _{\mathrm{e}} \mathrm{x}\right] } & =1 ; \mathrm{x} \in[1,2) \end{aligned} $

$ \therefore \mathrm{f}(\mathrm{x})=\left\{\begin{array}{l} \mathrm{e}^{\mathrm{x}^2} ; \mathrm{x} \in[0,1) \\\\ \mathrm{e} ; \mathrm{x} \in[1,2) \end{array}\right. $

$ \begin{aligned} & \int\limits_0^2 x f(x)=\int\limits_0^1 x \cdot e^{x^2} d x+\int\limits_1^2 x \cdot e d x \\\\ & x^2=t \Rightarrow 2 x d x=d t \end{aligned} $

$ =\frac{1}{2} \int\limits_0^1 e^t d t+\left.e \frac{x^2}{2}\right|_1 ^2 $

$ \begin{aligned} & =\frac{1}{2}(\mathrm{e}-1)+\frac{1}{2}(4-1) \mathrm{e} \\\\ & =2 \mathrm{e}-\frac{1}{2} \end{aligned} $

$\int_\limits{-\log _{e} 2}^{\log _{e} 2} e^{x}\left(\log _{e}\left(e^{x}+\sqrt{1+e^{2 x}}\right)\right) d x$

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

When, $x \rightarrow-\log _e 2$, then $t \rightarrow \frac{1}{2}$

When, $x \rightarrow \log _e 2$, then $t \rightarrow 2$

$ I=\int_\limits{\frac{1}{2}}^2\left[\log _e\left(t+\sqrt{1+t^2}\right)\right] d t $ ...........(i)

On applying integration by part method in Eq. (i), we get

$ \begin{aligned} & I=\left[t \log _e\left(t+\sqrt{1+t^2}\right)\right]_{1 / 2}^2-\int_{1 / 2}^2 \frac{t}{t+\sqrt{1+t^2}}\left(1+\frac{2 t}{2 \sqrt{1+t^2}}\right) d t \\\\ & =2 \log _e(2+\sqrt{5})-\frac{1}{2} \log _e\left(\frac{1+\sqrt{5}}{2}\right)-\int_{1 / 2}^2 \frac{t}{\sqrt{1+t^2}} d t \end{aligned} $

$ =\log _e\left(\frac{(2+\sqrt{5})^2}{\left(\frac{1+\sqrt{5}}{2}\right)^{1 / 2}}\right)-\frac{1}{2} \int_{1 / 2}^2 \frac{2 t}{\sqrt{1+t^2}} d t $ .............(ii)

Let $\quad I_1=\int_{1 / 2}^2 \frac{2 t}{\sqrt{1+t^2}} d t$

Let $1+t^2=w$

$2 t d t=d w$

When, $t \rightarrow \frac{1}{2}$, then $w=\frac{5}{4}$

When, $t \rightarrow 2$, then $w=5$

$ \begin{aligned} I_1 & =\int_{5 / 4}^5 \frac{1}{\sqrt{w}} d w \\\\ & =[2 \sqrt{w}]_{5 / 4}^5 \\\\ & =2\left[\sqrt{5}-\frac{\sqrt{5}}{2}\right]=\sqrt{5} \end{aligned} $

On substitute value of $I_1$ in Eq. (ii), we get

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

Given that

$ \int\limits_0^{t^2}\left(f(x)+x^2\right) d x=\frac{4}{3} t^3, \forall t>0 $

On differentiating using Leibnitz rule, we get

$ \begin{aligned} & \left(f\left(t^2\right)+t^4\right) \times 2 t=\frac{4}{3} \times 3 t^2 \\\\ & \Rightarrow f\left(t^2\right)+t^4=2 t \\\\ & \Rightarrow f\left(t^2\right)=2 t-t^4 \end{aligned} $

On substituting $\frac{\pi}{2}$ for $t$, we get

$ f\left(\frac{\pi^2}{4}\right)=2\left(\frac{\pi}{2}\right)-\frac{\pi^4}{16}=\pi\left(1-\frac{\pi^3}{16}\right) $

Let $I=\int\limits_0^\pi f(x) \sin x d x$ ..........(i)

$ \begin{aligned} & =\int\limits_0^\pi f(\pi-x) \sin (\pi-x) d x \\\\ & =\int\limits_0^\pi f(\pi-x) \sin x d x ........(ii) \end{aligned} $

On adding Equations (i) and (ii), we get

$ \begin{aligned} & 2 I=\int\limits_0^\pi[f(x)+f(\pi-x)] \sin x d x \\\\ & \Rightarrow 2 I=\int\limits_0^\pi \pi^2 \sin x d x=\pi^2 \int\limits_0^\pi \sin x d x \\\\ & =\pi^2(-\cos x)_0^\pi=-\pi^2(-1-1)=2 \pi^2 \\\\ & \Rightarrow I=\pi^2 \end{aligned} $

Let $L=\lim\limits_{n \rightarrow \infty}\left\{\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)\left(2^{\frac{1}{2}}-2^{\frac{1}{5}}\right) \ldots\left(2^{\frac{1}{2}}-2^{\frac{1}{2 n-1}}\right)\right\}$

Here, $\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)^n<\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)\left(2^{\frac{1}{3}}-2^{\frac{1}{5}}\right)$ $ \ldots\left(2^{\frac{1}{2}}-2^{\frac{1}{2 n-1}}\right)<\left(2^{\frac{1}{2}}-2^{\frac{1}{2 n-1}}\right)^n $

$ \Rightarrow \lim\limits_{n \rightarrow \infty}\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)^n < L < \lim\limits_{n \rightarrow \infty}\left(2^{\frac{1}{2}}-2^{\frac{1}{2 n-1}}\right)^n $

Since, $\lim\limits_{n \rightarrow \infty}\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)^n=0$ and $\lim\limits_{n \rightarrow \infty}\left(2^{\frac{1}{2}}-2^{\frac{1}{2 n-1}}\right)^n=0$

$\therefore L = 0$

We have,

$ 5 f(x)+4 f\left(\frac{1}{x}\right)=\frac{1}{x}+3, x>0 $ ..........(i)

On replacing $x$ by $\frac{1}{x}$ in (i), we get

$ 5 f\left(\frac{1}{x}\right)+4 f(x)=x+3 $ ..........(ii)

Now, using Eq. (i) $\times 5-$ (ii) $\times 4$, we get

$ \begin{aligned} & 25 f(x)-16 f(x) =\left(\frac{5}{x}+15\right)-(4 x+12) \\\\ &\Rightarrow 9 f(x) =\frac{5}{x}-4 x+3 \\\\ &\Rightarrow f(x) =\frac{1}{9}\left(\frac{5}{x}-4 x+3\right) ...........(iii) \end{aligned} $

$ \begin{aligned} \therefore \quad & 18 \int_1^2 f(x) d x=18 \int_1^2 \frac{1}{9}\left(\frac{5}{x}-4 x+3\right) d x \text { [Using Eq. (iii)] }\\\\ = & 2 \int_1^2\left(\frac{5}{x}-4 x+3\right) d x \end{aligned} $

$ \begin{aligned} & =2\left[5 \log _e x-4\left(\frac{x^2}{2}\right)+3 x\right]_1^2 \\\\ & =2\left[\left(5 \log _e 2-2(2)^2+3(2)\right)-\left(5 \log 1-2(1)^2+3(1)\right)\right] \\\\ & =2\left[5 \log _e 2-8+6+2-3\right] [\because \log 1=0]\\\\ & =10 \log _e 2-6 \end{aligned} $

$ I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{x+\frac{\pi}{4}}{(2-\cos 2 x)} d x $

Using $\int_a^b f(x) d x=\int_a^b f(a+b-x) d x$

$ I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left(\frac{-x+\frac{\pi}{4}}{2-\cos 2 x}\right) d x $

$\begin{aligned} & \therefore 2 I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\pi d x}{2(2-\cos 2 x)} \\\\ & \Rightarrow I=\frac{2 \pi}{4} \int_0^{\frac{\pi}{4}}\left(\frac{d x}{\left.\frac{2-1-\tan ^2 x}{1+\tan ^2 x}\right)}\right. \\\\ & \Rightarrow I=\frac{\pi}{2} \int_0^{\frac{\pi}{4}}\left(\frac{1+\tan ^2 x}{1+3 \tan ^2 x}\right) d x\end{aligned}$

Now,

$ \begin{aligned} & \tan x=t \\\\ & =\frac{\pi}{2} \int_0^1 \frac{d t}{1+3 t^2} \\\\ & =\frac{\pi}{2}\left[\frac{\tan ^{-1}(\sqrt{3} t)}{\sqrt{3}}\right]_0^1 \\\\ & =\frac{\pi}{2 \sqrt{3}}\left(\frac{\pi}{3}\right)=\frac{\pi^2}{6 \sqrt{3}} \end{aligned} $

$ \begin{aligned} & \lim _{n \rightarrow \infty}\left[\frac{1}{1+n}+\frac{1}{2+n}+\frac{1}{3+n}+\ldots \ldots+\frac{1}{2 n}\right] \\\\ & =\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{1}{r+n} \\\\ & =\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{1}{n}\left(\frac{1}{\frac{r}{n}+1}\right) \\\\ & =\int_0^1 \frac{d x}{x+1} \\\\ & =\left.\log _e(1+\mathrm{x})\right|_0 ^1 \\\\ & =\log _e^2 \end{aligned} $

$I=\int\limits_{0}^{\alpha} \frac{x}{\sqrt{x+\alpha}-\sqrt{x}} d x, \alpha>0$

$ \begin{aligned} & =\frac{1}{\alpha} \int\limits_{0}^{\alpha} x(\sqrt{x+\alpha}+\sqrt{x}) d x \end{aligned} $

$\eqalign{ & = {1 \over \alpha }\int\limits_0^\alpha {\left( {x\sqrt {x + \alpha } + x\sqrt x } \right)} dx \cr & = {1 \over \alpha }\int\limits_0^\alpha {\left[ {\left( {x + \alpha - \alpha } \right)\sqrt {x + \alpha } + {x^{{3 \over 2}}}} \right]} dx \cr} $

$ \begin{aligned} & =\frac{1}{\alpha}\int\limits_0^\alpha \left[(x+\alpha)^{3 / 2}-\alpha(x+\alpha)^{1 / 2}+x^{3 / 2}\right] d x \\\\ & =\frac{1}{\alpha}\left[\frac{2}{5}(x+\alpha)^{5 / 2}-\alpha \frac{2}{3}(x+\alpha)^{3 / 2}+\frac{2}{5} x^{5 / 2}\right]_0^\alpha \\\\ & =\frac{1}{\alpha}\left(\frac{2}{5}(2 \alpha)^{5 / 2}-\frac{2 \alpha}{3}(2 \alpha)^{3 / 2}+\frac{2}{5} \alpha^{5 / 2}-\frac{2}{5} \alpha^{5 / 2}+\frac{2}{3} \alpha^{5 / 2}\right) \\\\ & =\frac{1}{\alpha}\left(\frac{2^{7 / 2} \alpha^{5 / 2}}{5}-\frac{2^{5 / 2} \alpha^{5 / 2}}{3}+\frac{2}{3} \alpha^{5 / 2}\right)=\alpha^{3 / 2}\left(\frac{2^{7 / 2}}{5}-\frac{2^{5 / 2}}{3}+\frac{2}{3}\right) \\\\ & =\frac{\alpha^{3 / 2}}{15}(24 \sqrt{2}-20 \sqrt{2}+10)=\frac{\alpha^{3 / 2}}{15}(4 \sqrt{2}+10) \end{aligned} $

Now, $\frac{\alpha^{3 / 2}}{15}(4 \sqrt{2}+10)=\frac{16+20 \sqrt{2}}{15}=\frac{2 \sqrt{2}(4 \sqrt{2}+10)}{15}$

$ \Rightarrow \alpha^{3 / 2}=2 \sqrt{2}=2^{3 / 2} $

$\Rightarrow \alpha=2 $

$\phi(x)=\frac{1}{\sqrt{x}} \int_{\pi / 4}^{x}\left(4 \sqrt{2} \sin t-3 \phi^{\prime}(t)\right) d t$

$\Rightarrow \phi^{\prime}(x)=\frac{-1}{2 x^{3 / 2}} \int_{\pi / 4}^{x}\left(4 \sqrt{2} \sin t-3 \phi^{\prime}(t)\right) d t$

$+\frac{1}{\sqrt{x}}\left(4 \sqrt{2} \sin (x)-3 \phi^{\prime}(x)\right)$

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

$\phi^{\prime}\left(\frac{\pi}{4}\right)=\frac{-1}{2\left(\frac{\pi}{4}\right)^{3 / 2}} \times 0+\sqrt{\frac{4}{\pi}}\left(4 \sqrt{2} \times \frac{1}{\sqrt{2}}-3 \phi^{\prime}\left(\frac{\pi}{4}\right)\right)$

$\Rightarrow \phi^{\prime}\left(\frac{\pi}{4}\right)\left[1+\frac{6}{\sqrt{\pi}}\right]=\frac{2}{\sqrt{\pi}} \times 4$

$\Rightarrow \phi^{\prime}\left(\frac{\pi}{4}\right)=\frac{8}{6+\sqrt{x}}$

$\int_{0}^{\alpha} \frac{t^{50}}{1-t} d t$

$ \begin{aligned} & = \int_{0}^{\alpha} \frac{t^{50}-1+1}{1-t}\\\\ & =-\int_{0}^{\alpha}\left(\frac{1-t^{50}}{1-t}-\frac{1}{1-t}\right) d t\\\\ & =-\int_{0}^{\alpha}\left(1+t+\ldots . .+t^{49}\right)+\int_{0}^{\alpha} \frac{1}{1-t} d t \end{aligned} $

$=-\left(\frac{\alpha^{50}}{50}+\frac{\alpha^{49}}{49}+\ldots . .+\frac{\alpha^{1}}{1}\right)+\left(\frac{\ln (1-\mathrm{f})}{-1}\right)_{0}^{\alpha}$

$=-\mathrm{P}_{50}(\alpha)-\ln (1-\alpha)$

$=-\mathrm{P}_{50}(\alpha)-\beta$

Let I = $\int_\limits{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{(2+3 \sin x)}{\sin x(1+\cos x)} d x$

$ =\int\limits_{\pi / 3}^{\pi / 2} \frac{2}{\sin x(1+\cos x)} d x+\int\limits_{\pi / 3}^{\pi / 2} \frac{3}{1+\cos x} d x $

$ \begin{aligned} =\int\limits_{\pi / 3}^{\pi / 2} \frac{2(1-\cos x)}{\sin x\left(1-\cos ^2 x\right)} & d x +\int\limits_{\pi / 3}^{\pi / 2} \frac{3}{1+\cos x} d x \end{aligned} $

$ =\int\limits_{\pi / 3}^{\pi / 2} \frac{2(1-\cos x)}{\sin ^3 x} d x+\int\limits_{\pi / 3}^{\pi / 2} \frac{3}{2 \cos ^2 \frac{x}{2}} d x $

$ \begin{array}{r} =\int\limits_{\pi / 3}^{\pi / 2} \frac{2}{\sin ^3 x} d x-\int\limits_{\pi / 3}^{\pi / 2} 2 \cot x \operatorname{cosec}^2 x d x +\int\limits_{\pi / 3}^{\pi / 2} \frac{3}{2} \sec ^2 \frac{x}{2} d x \end{array} $

$ =2 I_1-2 I_2+\frac{3}{2} I_3 $

Now,

$ \begin{aligned} I_1 & =\int\limits_{\pi / 3}^{\pi / 2} \operatorname{cosec}^3 x d x \\\\ & =\int\limits_{\pi / 3}^{\pi / 2} \sqrt{1+\cot ^2 x} \operatorname{cosec}^2 x d x \end{aligned} $

Put $\cot x=t \Rightarrow-\operatorname{cosec}^2 x d x=d t$

When, $x=\frac{\pi}{3} \Rightarrow t=\frac{1}{\sqrt{3}} \text { and } x=\frac{\pi}{2} \Rightarrow t=0$

$ \begin{aligned} \therefore & I_1=-\int\limits_{\frac{1}{\sqrt{3}}}^0 \sqrt{1+t^2} d t \\\\ = & \int\limits_0^{\frac{1}{\sqrt{3}}} \sqrt{1+t^2} d t \\\\ = & {\left[\frac{t}{2} \sqrt{1+t^2}+\frac{1}{2} \log \left[t+\sqrt{1+t^2}\right]\right]_0^{\frac{1}{\sqrt{3}}} } \\\\ = & \frac{1}{3}+\frac{1}{2} \log \sqrt{3} \end{aligned} $

$ I_2=\int\limits_{\pi / 3}^{\pi / 2} \cot x \operatorname{cosec}^2 x d x $

Put $\cot x=t \Rightarrow \operatorname{cosec}^2 x d x=-d t$

When, $x=\frac{\pi}{3}$ $\Rightarrow t=\frac{1}{\sqrt{3}}$ and $x=\frac{\pi}{2} \Rightarrow t=0$

$ \therefore \begin{aligned} I_2 & =-\int\limits_{\frac{1}{\sqrt{3}}}^0 t d t \\\\ & =\int\limits_0^{\frac{1}{\sqrt{3}} t} t d t=\left[\frac{t^2}{2}\right]_0^{\frac{1}{\sqrt{3}}}=\frac{1}{6} \end{aligned} $

$ \begin{aligned} & I_3=\int\limits_{\frac{\pi}{3}}^{\frac{\pi}{2}} \sec ^2 \frac{x}{2} d x \\\\ = & 2\left(\tan \frac{\pi}{4}-\tan \frac{\pi}{6}\right)=2\left(1-\frac{1}{\sqrt{3}}\right) \end{aligned} $

$ \begin{aligned} & \therefore \quad I=2 I_1-2 I_2+\frac{3}{2} I_3 \\\\ & =2\left(\frac{1}{3}+\frac{1}{2} \log \sqrt{3}\right)-2\left(\frac{1}{6}\right) +\frac{3}{2}\left(2\left(1-\frac{1}{\sqrt{3}}\right)\right) \end{aligned} $

$ \begin{aligned} & =\frac{2}{3}+\log \sqrt{3}-\frac{1}{3}+3-\sqrt{3} \\\\ & =\frac{10}{3}-\sqrt{3}+\log _e \sqrt{3} \end{aligned} $

$\mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 0}^{n - 1} {{3 \over n}{{\left( {2 + {r \over n}} \right)}^2}} $

$ = \int_0^1 {3{{(2 + x)}^2}\,dx} $

$ = \left. {3\,.\,{{{{(2 + x)}^3}} \over 3}} \right|_0^1$

$ = {3^3} - {2^3} = 19$

$I = {{3(e - 1)} \over e}\int_1^2 {{x^2}{e^{[x] + [{x^3}]}}dx} $

$ = {{3(e - 1)} \over e}\int_1^2 {{x^2}{e^{1 + [{x^3}]}}dx} $ ($\because$ $[x] = 1$ when $x \in (12)$)

$ = 3(e - 1)\int_1^2 {{x^2}{e^{[{x^3}]}}dx} $

Let ${x^3} = t$

$I = (e - 1)\int_1^8 {{e^{[t]}}dt} $

$ = ({e^{ - 1}})({e^1} + {e^2} + {e^3}\, + \,...\, + \,{e^7})$

$ = (e - 1)e{{({e^7} - 1)} \over {e - 1}}$

$ = {e^8} - e$