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.

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

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)

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

$ \int_{\frac{\pi}{5}}^{\frac{3 \pi}{10}} \frac{d x}{\sec ^{2} x+\left(\tan ^{2024} x-1\right)\left(\sec ^{2} x-1\right)} $

$ \begin{array}{l}\int_{\frac{2 \pi}{10}}^{\frac{3 \pi}{0}} \frac{d x}{1+\tan ^{2} x+\tan ^{2024} x-\tan ^{2} x} \\\\ =\int_{\frac{2 \pi}{10}}^{\frac{3 \pi}{10}} \frac{d x}{1+\tan ^{2024} x} \\\\ I=\int_{\frac{2 \pi}{10}}^{\frac{3 \pi}{0}} \frac{\cos ^{2024} x}{\sin ^{2024} x+\cos ^{2024} x} d x \end{array} $

Applyh P_IV

$ I=\int_{\frac{\pi}{10}}^{\frac{3 \pi}{10}} \frac{\sin ^{2024} x}{\cos ^{2024} x+\sin ^{2024} x} d x $

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

$ \begin{array}{l} 2 I=\int_{\frac{\pi}{5}}^{\frac{3 \pi}{10}} d x=(x)_{\pi / 5}^{3 \pi / 10}=\frac{3 \pi}{10}-\frac{2 \pi}{10}=\frac{\pi}{10} \\\\ I=\frac{\pi}{20} \end{array} $

Apply P - IV

$ \begin{aligned} I & =\int_{\frac{-\pi}{15}}^{\frac{\pi}{15}} \frac{\cos (-5 x)}{1+e^{-5 x}} d x \\\\ & =\int_{\frac{-\pi}{15}}^{\frac{\pi}{15}} \frac{e^{5 x} \cos 5 x}{e^{5 x}+1} d x \end{aligned} $

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

$ \begin{array}{l} 2 I=\int_{\frac{-\pi}{15}}^{\frac{\pi}{15}} \frac{\cos 5 x\left(e^{5 x}+1\right)}{e^{5 x}+1} d x=\int_{\frac{-\pi}{15}}^{\frac{\pi}{15}} \cos 5 x d x \\\\ =2 \int_{0}^{\frac{\pi}{15}} \cos 5 x d x \\\\ 2 I=2 \frac{1}{5}[\sin 5 x]_{0}^{\frac{\pi}{15}} \\\\ I=\frac{1}{5}\left[\sin \frac{\pi}{3}-0\right]=\frac{1}{5} \times \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{10} \end{array} $

To solve the given integral problem, we can start by simplifying the expression inside the integral:

$ \frac{3}{25} \int_{0}^{25 \pi} \sqrt{\left|\cos x - \cos^3 x\right|} \, dx $

We can rewrite this expression as:

$ \frac{3}{25} \int_{0}^{25 \pi} \sqrt{|\cos x (1 - \cos^2 x)|} \, dx $

Recognizing that $1 - \cos^2 x = \sin^2 x$, this becomes:

$ \frac{3}{25} \int_{0}^{25 \pi} \sqrt{|\cos x| \sin^2 x} \, dx $

We further simplify this to:

$ \frac{3}{25} \int_{0}^{25 \pi} |\sin x| \sqrt{|\cos x|} \, dx $

Next, we decompose the integral over the period from $0$ to $25\pi$ into parts over smaller intervals where the function symmetry and periodicity can be utilized:

$ = \frac{3}{25} \left[13 \int_{0}^{\frac{\pi}{2}} \sin x \sqrt{\cos x} \, dx + 13 \int_{\frac{\pi}{2}}^{\pi} \sin x \sqrt{-\cos x} \, dx - 12 \int_{\pi}^{\frac{3\pi}{2}} \sin x \sqrt{-\cos x} \, dx - 12 \int_{\frac{3\pi}{2}}^{2\pi} \sin x \sqrt{\cos x} \, dx\right] $

For each sub-interval, we compute the integral:

For $ \int_{0}^{\frac{\pi}{2}}$, use the substitution $\cos x$ from 1 to 0.

For $ \int_{\frac{\pi}{2}}^{\pi}$, and similarly mapped intervals $\int_{\pi}^{\frac{3\pi}{2}}$, $\int_{\frac{3\pi}{2}}^{2\pi}$, follow similar transformations respecting their respective limits to evaluate each segment.

The integrals resolve as follows:

$ = \frac{3}{25} \left[-13 \left[\frac{2(\cos x)^{\frac{3}{2}}}{3} \right]_{0}^{\frac{\pi}{2}} + 13 \left[\frac{2(-\cos x)^{\frac{3}{2}}}{3}\right]_{\frac{\pi}{2}}^{\pi} - 12\left[\frac{2(-\cos x)^{\frac{3}{2}}}{3}\right]_{\pi}^{\frac{3\pi}{2}} + 12 \left[\frac{2}{3}(\cos x)^{\frac{3}{2}} \right]_{\frac{3\pi}{2}}^{2\pi}\right] $

Evaluating each integral gives:

$ = \frac{3}{25} \times \frac{1}{3}\left[-26(0-1) + 26(1-0) - 24(0-1) + 24(1-0)\right] $

This simplifies to:

$ = \frac{1}{25}[26 + 26 + 48] = \frac{1}{25} \times 100 = 4 $

Thus, the final calculated value of the integral is 4.

To analyze the integrals, we start by considering the expression $ I = \int_{-2\pi}^{2\pi} \sin^m x \cos^n x \, dx $.

Since the given condition is:

$ I = 4 \int_{0}^{\pi} \sin^m x \cos^n x \, dx $

We know that:

$ \int_{-a}^{a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx $

if $ f(x) $ is an even function. This suggests that $ \sin^m x \cos^n x $ might be even in this case. To satisfy the condition where the integral over the larger symmetric interval is four times the integral over the smaller positive interval, $ m $ must be even.

Similarly, consider:

$ \int_{-\pi}^{\pi} \sin^r x \cos^s x \, dx = 4 \int_{0}^{\pi/2} \sin^r x \cos^s x \, dx $

Here, for a comparable reason, we must have $ r $ as even.

Finally, for:

$ \int_{-\pi/2}^{\pi/2} \sin^l x \cos^m x \, dx = 0 $

the function should be odd because the integral of an odd function over a symmetric interval around zero results in zero. Thus, $ l $ must be odd.

With the constraints $ 9 > m > l > s > n > r > 2 $, and that $ m $ and $ r $ are even while $ l $ is odd, a possible assignment of these integers could be:

$ m = 8 $

$ l = 7 $

$ s = 6 $

$ n = 5 $

$ r = 4 $

Next, let's test the option that matches this setup:

For option C, $(s-2)(s+2) = ln - 3$:

$ (s-2)(s+2) = (6-2)(6+2) = 4 \times 8 = 32 $

The expression $ ln - 3 $ was a typo originally intended to test our guess, but $ 32 = 32 $ checks out, confirming this is our desired setup.

Given, $ I=\int_{0}^{\pi}\left(\sin ^{3} x+\cos ^{2} x\right)^{2} d x $

$ \begin{array}{c} I=\int_{0}^{\pi}\left[\sin ^{3}(\pi-x)+\cos ^{2}(\pi-x)\right]^{2} d x \\\\ I=\int_{0}^{\pi}\left(\sin ^{3} x+\cos ^{2} x\right)^{2} d x \\\\ \Rightarrow I=2 \int_{0}^{\pi / 2}\left(\sin ^{3} x+\cos ^{2} x\right)^{2} d x \end{array} $

By applying the property,

$ I=2 \int_{0}^{\pi / 2}\left(\cos ^{3} x+\sin ^{2} x\right)^{2} d x $

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

$ \begin{array}{c} 2 I=2\left[\int _ { 0 } ^ { \pi / 2 } \left(\left(\sin ^{3} x+\cos ^{2} x\right)^{2}\right.\right. \\\\ \left.+\left(\cos ^{3} x+\sin ^{2} x\right)^{2}\right] d x \\\\ I=\int_{0}^{\pi / 2} \sin ^{6} x+\cos ^{6} x+\sin ^{4} x+\cos ^{4} x \\\\ +2 \sin ^{3} x \cos ^{2} x+2 \cos ^{3} x \sin ^{2} x d x \\\\ I=\int_{0}^{\pi / 2}\left(\sin ^{2} x+\cos ^{2} x\right)\left(\sin ^{4} x\right. \\\\ \left.+\cos ^{4} x-\sin ^{2} x \cos ^{2} x\right) \\\\ +\sin ^{4} x+\cos ^{4} x+2 \sin ^{2} x \cos ^{2} \end{array} $

$ x(\sin x+\cos x) d x $

$ I=\int_{0}^{\pi / 2} 2\left(\sin ^{4} x+\cos ^{4} x\right)+\sin ^{2} x \cos ^{2} $

$ x\left(2 \sin x+2 \cos x^{-1}\right) $

$ I=\int_{0}^{\pi / 2} 2\left(1-2 \sin ^{2} x \cos ^{2} x\right)+\sin ^{2} x \cos ^{2} $

$ x(2 \sin x+2 \cos x-1) d x $

$ \begin{array}{r} I=\int_{0}^{\pi / 2} 2-5 \sin ^{2} x \cos ^{2} x+2 \sin ^{3} x \cos ^{2} \\\\ x+2 \sin ^{2} x \cos ^{3} x d x \end{array} $

Let $ I=\int_{0}^{\pi / 2} 5 \sin ^{2} x \cos ^{2} x d x $

$ =\frac{5}{4} \int_{0}^{\pi / 2} \sin ^{2} 2 x d x $

$ =\frac{5}{4} \int_{0}^{\pi / 2} \frac{1-\cos 4 x}{2} d x $

$ =\frac{5}{8}\left(x-\frac{\sin 4 x}{4}\right)_{0}^{\pi / 2}=\frac{5}{8}\left[\frac{\pi}{2}\right]=\frac{5 \pi}{16} $

$ I_{2}=\int_{0}^{\pi / 2} 2 \sin ^{3} x \cos ^{2} x d x $

$ =\int_{0}^{\pi / 2}\left(1-\cos ^{2} x\right) \cos ^{2} x \sin x d x $

Let $ \cos x=t $

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

Also, $ x=0, t=1 $

$ \begin{array}{l} x=\pi / 2, t=0 \\\\ \begin{aligned} I_{2} & =2 \int_{1}^{0}\left(t^{2}-1\right) t^{2} d t=2 \int_{1}^{0}\left(t^{4}-t^{2}\right) d t \\\\ & =2\left[\frac{t^{5}}{5}-\frac{t^{3}}{3}\right]_{1}^{0}=2\left[-\frac{1}{5}+\frac{1}{3}\right]=\frac{4}{15} \\\\ I_{3} & =2 I=\int_{0}^{\pi / 2} \sin ^{2} x \cos ^{3} x d x \\\\ I_{3} & =2 \int_{0}^{\pi / 2} \sin ^{2} x\left(1-\sin ^{2} x\right) \cos x d x \end{aligned} \end{array} $

Let $ \sin x=u \Rightarrow \cos x d x=d u $

Also, $ x=0, u=0 \Rightarrow x=\pi / 2, u=1 $

$ \begin{aligned} I_{3} & =2 \int_{0}^{1} u^{2}\left(1-u^{2}\right) d u \\\\ & =2\left[\frac{u^{3}}{3}-\frac{u^{5}}{5}\right]_{0}^{1}=2\left[\frac{1}{3}-\frac{1}{5}\right]=\frac{4}{15} \\\\ I & =[2 x]_{0}^{\pi / 2}-\frac{5 \pi}{16}+\frac{4}{15}+\frac{4}{15} \\\\ & =\pi-\frac{5 \pi}{16}+\frac{8}{15}=\frac{11 \pi}{16}+\frac{8}{15} \end{aligned} $

Given, $ I=\int_{-\frac{\pi}{8}}^{\pi / 8} \frac{\sin ^{4}(4 x)}{1+e^{4 x}} d x $

$ \begin{array}{l} I=\int_{0}^{\pi / 8}\left[\frac{\sin ^{4}(4 x)}{1+e^{4 x}}+\frac{\sin ^{4}(-4 x)}{1+e^{-4 x}}\right] d x \\\\ I=\int_{0}^{\pi / 8} \sin ^{4}(4 x)\left[\frac{1}{1+e^{4 x}}+\frac{e^{4 x}}{e^{4 x}+1}\right] d x \\\\ I=\int_{0}^{\pi / 8} \sin ^{4} 4 x d x \end{array} $

Let $ 4 x=t \Rightarrow 4 d x=d t $

If $ x=0 \Rightarrow t=0 $

$ \begin{array}{c} x=\frac{\pi}{8} \Rightarrow t=\frac{\pi}{2} \\\\ I=\frac{1}{4} \int_{0}^{\pi / 2} \sin ^{4} t d t \\\\ I=\frac{1}{4} \int_{0}^{\pi / 2} \sin ^{4}\left(\frac{\pi}{2}-t\right) \\\\ =\frac{1}{4} \int_{0}^{\pi / 2} \cos ^{4} t d t \end{array} $

$ \begin{array}{l} \quad I=\frac{1}{4} \int_{0}^{\pi / 2} \sin ^{4}\left(\frac{\pi}{2}-t\right) d t \\\\ \quad=\frac{1}{4} \int_{0}^{\pi / 2} \cos ^{4} t d t \end{array} $

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

$ \begin{array}{l} 2 I=\frac{1}{4} \int_{0}^{\pi / 2}\left(\sin ^{4} t+\cos ^{4} t\right) d t \\\\ I=\frac{1}{8} \int_{0}^{\pi / 2}\left[1-\frac{1}{2}(\sin 2 t)^{2}\right] d t \\\\ I=\frac{1}{8} \int_{0}^{\pi / 2}\left[1-\frac{1}{4}(1-\cos 4 t)\right] d t \\\\ I=\frac{1}{8} \int_{0}^{\pi / 2}\left[\frac{3}{4}+\frac{1}{4} \cos 4 t\right] d t \\\\ I=\frac{1}{8}\left(\frac{3 t}{4}+\frac{\sin 4 t}{16}\right)_{0}^{\pi / 2} \Rightarrow I=\frac{1}{8}\left(\frac{3 \pi}{8}\right)=\frac{3 \pi}{64} \end{array} $

Let $I=\int_{-\frac{3}{4}}^{\frac{\pi-6}{8}} \log (\sin (4 x+3)) d x \quad \ldots \text { (i) } $

$ \begin{aligned} & \Rightarrow I=\int_{-\frac{3}{4}}^{\frac{\pi-6}{8}} \log \left(\sin \left(4\left(\frac{\pi-6}{8}+\left(-\frac{3}{4}\right)-x\right)+3\right)\right) d x \\\\ & \Rightarrow I=\int_{-\frac{3}{4}}^{\frac{\pi-6}{8}} \log \left(\sin \left(\frac{\pi}{2}-(4 x+3)\right)\right) d x \\\\ & \quad \Rightarrow I=\int_{-\frac{3}{4}}^{\frac{\pi-6}{8}} \log (\cos (4 x+3)) d x \quad \ldots \text { (ii) } \end{aligned} $

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

$ \begin{aligned} & 2 I=\int_{-\frac{3}{4}}^{\frac{\pi-6}{8}} \log (\sin (4 x+3) \cos (4 x+3)) d x \\\\ & \Rightarrow 2 I=\int_{-\frac{3}{4}}^{\frac{\pi-6}{8}} \log \left(\frac{\sin (2(4 x+3))}{2}\right) d x \end{aligned} $

$ \begin{aligned} & \Rightarrow 2 I=\int_{-\frac{3}{4}}^{\frac{\pi-6}{8}} \log (\sin (2(4 x+3))) d x \\\\ & \qquad-\int_{-\frac{3}{4}}^{\frac{\pi-6}{8}}(\log 2) d x \\\\ & \Rightarrow 2 I=I-(\log 2)\left[\frac{\pi-6}{8}-\left(-\frac{3}{4}\right)\right] \\\\ & \Rightarrow I=-\frac{\pi}{8} \log 2 \end{aligned} $

$ \begin{aligned} & \text {Let } I=\int_0^{16} \frac{\sqrt{x}}{1+\sqrt{x}} d x \\\\ & \text { Put } 1+\sqrt{x}=t \\\\ & \Rightarrow \quad \frac{1}{2 \sqrt{x}} d x=d t \\\\ & \Rightarrow \quad d x=2(t-1) d t \\\\ & \text { When } x=0, t=1 \\\\ & \text { When } x=16, t=5 \\\\ & \begin{aligned} \therefore I & =\int_1^{5 t-1} \\\\ t & (2(t-1)) d t \\\\ & =2 \int_1^{5 t^2-2 t+1} \\\\ & =2 \int_1^5\left(t-2+\frac{1}{t}\right) d t \\\\ & =2\left[\frac{t^2}{2}-2 t+\log t\right]_1^5 \end{aligned} \end{aligned} $

$ \begin{aligned} & =2\left[\left\{\frac{25}{2}-10+\log 5\right\}-\left\{\frac{1}{2}-2+\log 1\right\}\right] \\\\ & =2\left[\frac{5}{2}+\log 5+\frac{3}{2}+0\right] \\\\ & =8+2 \log 5 \end{aligned} $

$ \begin{aligned} & \text { Let } I=\int_0^{32 \pi} \sqrt{1-\cos 4 x} d x \\\\ & =\int_0^{32 \pi} \sqrt{1-\left(1-2 \sin ^2 2 x\right)} d x \\\\ & =\int_0^{32 \pi} \sqrt{2}|\sin 2 x| d x \end{aligned} $

The period of $|\sin 2 x|$ is $\frac{\pi}{2}$

$ \begin{aligned} \therefore \quad I & =64 \int_0^{\pi / 2} \sqrt{2} \sin 2 x d x \\\\ & =64 \sqrt{2}\left[\frac{-\cos 2 x}{2}\right]_0^{\pi / 2} \\\\ & =32 \sqrt{2}[-\cos \pi+\cos 0] \\\\ & =32 \sqrt{2}[-(-1)+1]=64 \sqrt{2} \end{aligned} $

We have,

$ \begin{aligned} f(x)= & \int \frac{\sin 2 x+2 \cos x}{4 \sin ^2 x+5 \sin x+1} d x \\\\ & =\int \frac{2 \cos x(\sin x+1)}{(4 \sin x+1)(\sin x+1)} d x \\\\ & =\int \frac{2 \cos x}{4 \sin x+1} d x \end{aligned} $

$\begin{aligned} & \text { Let } 4 \sin x+1=t \Rightarrow 4 \cos x d x=d t \\\\ & \begin{aligned} \Rightarrow & f(x)=\frac{1}{2} \int \frac{d t}{t}=\frac{1}{2} \log t+C \\\\ & =\frac{1}{2} \log (4 \sin x+1)+C\end{aligned} \\\\ & \begin{aligned} \because & f(0)=0 \Rightarrow C=0 \\\\ \therefore & f(x)=\frac{1}{2} \log (4 \sin x+1)\end{aligned} \\\\ & f\left(\frac{\pi}{6}\right)=\frac{1}{2} \log \left(\frac{4}{2}+1\right)=\frac{1}{2} \log 3\end{aligned}$

Let $I=\int_{-2}^2 x^4\left(4-x^2\right)^{7 / 2} d x$

$ \begin{aligned} & I=2 \int_0^2 x^4\left(4-x^2\right)^{7 / 2} d x \\\\ & \qquad \quad[\because f(x) \text { is even function }] \\\\ & \text { put } x=2 \sin \theta \Rightarrow d x=2 \cos \theta d \theta \\\\ & \text { when } x=0, \theta=0 \text { and } x=2, \theta=\frac{\pi}{2} \\\\ & \therefore I=2 \int_0^{\pi / 2} 2^4 \sin ^4 \theta(2 \cos \theta)^7 \cdot 2 \cos \theta d \theta \\\\ & I=2 \cdot 2^4 \cdot 2^8 \int_0^{\pi / 2} \sin ^4 \theta \cos ^8 \theta d \theta \\\\ & I=2^{13} . \end{aligned} $

$ \frac{(4-1)(4-3)(8-1)(8-3)(8-5)(8-7)}{12 \times 10 \times 8 \times 6 \times 4 \times 2} \times \frac{\pi}{2} $

(by Wall's formulae)

$ \begin{aligned} & =\frac{2^{13} \times 3 \times 1 \times 7 \times 5 \times 3 \times 1 \times \frac{\pi}{2}}{12 \times 10 \times 8 \times 6 \times 4 \times 2} \\\\ & =\frac{2^6 \times 3 \times 7 \times 5 \times 3 \times \pi}{6 \times 5 \times 4 \times 3 \times 2}=28 \pi \end{aligned} $

Let $I=\int_0^{2 \pi}\left(\sin ^4 x+\cos ^4 x\right) d x$

As $\sin ^4 x+\cos ^4 x$ is periodic with $\frac{\pi}{2}$

So, $\int_0^{2 \pi}\left(\sin ^4 x+\cos ^4 x\right) d x$

$ \begin{aligned} & =\int_0^{4 \times \frac{\pi}{2}}\left(\sin ^4 x+\cos ^4 x\right) d x \\ = & 4 \int_0^{\pi / 2}\left(\sin ^4 a x+\cos ^4 x\right) d x \\ = & 4 \int_0^{\pi / 2} 1-\frac{\sin ^2 2 x}{2} d x \\ = & 4 \int_0^{\pi / 2} \frac{4-1+\cos 4 x}{4} d x \\ = & 4 \int_0^{\pi / 2} \frac{3+\cos 4 x}{4} d x \\ = & {\left[3-\frac{\sin 4 x}{4}\right]_0^{\frac{\pi}{2}} } \\ = & 3\left[\frac{\pi}{2}\right]=\frac{3 \pi}{2} \end{aligned} $

Now, $\int_0^\pi \sin ^2 x d x=2 \int_0^{\frac{\pi}{2}} \sin ^2 x d x$

$ \begin{aligned} & =2 \int_0^{\frac{\pi}{2}}\left(\frac{1-\cos 2 x}{2}\right) d x=\left[x-\frac{\sin 2 x}{2}\right]_0^{\frac{\pi}{2}} \\ & =\frac{\pi}{2} \end{aligned} $

Similarly,

$ \begin{aligned} & \int_0^{\frac{\pi}{2}} \cos ^2 x d x=\frac{\pi}{4} \\ \therefore & \frac{3 \pi}{2}=K \times \frac{\pi}{2}+L \times \frac{\pi}{4} \\ & 2 K+L=6 \end{aligned} $

If $K=1$, then $L=4$ and if $K=2$, then $L=2$

$\therefore$ There are two possible ordered pair i.e. $(1,4),(2,2)$.

$ \text { } \begin{aligned}Let \,\, I & =\int_0^\pi \frac{x \sin x d x}{4 \cos ^2 x+3 \sin ^2 x} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & =\int_0^\pi \frac{x \sin x d x}{4 \cos ^2 x+3-3 \cos ^2 x} \\ & =\int_0^\pi \frac{x \sin x}{\cos ^2 x+3} \\ & =\int_0^\pi \frac{(\pi-x) \sin (\pi-x)}{\cos ^2(\pi-x)+3} \end{aligned} $

$ \begin{aligned} &=\int_0^\pi \frac{(\pi-x) \sin x}{\cos ^2 x+3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ &\text { On adding Eq. (i) and (ii), we get }\\ &\begin{aligned} 2 I & =\int_0^\pi \frac{\pi \sin x}{3+\cos ^2 x} d x \\ & =\pi \int_0^\pi \frac{\sin x}{3+\cos ^2 x} d x=-\pi \int_1^{-1} \frac{1}{3+t^2} d t \\ & =-\pi \frac{1}{\sqrt{3}}\left(\tan ^{-1} \frac{t}{\sqrt{3}}\right)_1^{-1} \\ \Rightarrow \quad 2 I & =\frac{-\pi}{\sqrt{3}}\left[\tan ^{-1}\left(-\frac{1}{\sqrt{3}}\right)-\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)\right] \\ & =\frac{-\pi}{\sqrt{3}}\left[-\frac{\pi}{6}-\frac{\pi}{6}\right] \\ & =-\frac{\pi}{\sqrt{3}} \times\left(-\frac{2 \pi}{6}\right)=\frac{\pi^2}{3 \sqrt{3}} \\ \Rightarrow \quad I & =\frac{\pi^2}{6 \sqrt{3}} \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { Let } I=\int\left(\frac{1+x^2}{1+x^4}\right) d x \\ & =\int \frac{\left(1+\frac{1}{x^2}\right)}{\left(x^2+\frac{1}{x^2}-2+2\right)} d x \\ & =\int \frac{1+\frac{1}{x^2}}{\left(x-\frac{1}{x}\right)^2+2} d x \end{aligned} $

Let $x-\frac{1}{x}=t$

$ \begin{aligned} & \Rightarrow\left(1+\frac{1}{x^2}\right) d x=d t \\ & =\int \frac{1}{t^2+2} d t=\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{t}{\sqrt{2}}\right) \\ & =\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{x^2-1}{\sqrt{2} x}\right) \text { or } \frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{2}}\right) \end{aligned} $

$ \begin{aligned} A & =\int_0^{\infty} \frac{1+x^2}{1+x^4} d x=\left[\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{2}}\right)\right]_0^{\infty} \\ & =\frac{1}{\sqrt{2}}\left[\tan ^{-1} \infty-\tan ^{-1}(-\infty)\right] \\ & =\frac{1}{\sqrt{2}} \times 2 \times \frac{\pi}{2} \Rightarrow \frac{\pi}{\sqrt{2}} \\ B & =\int_0^1 \frac{1+x^2}{1+x^4} d x=\left[\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{2}}\right)\right]_0^1 \\ & =\frac{1}{\sqrt{2}}\left[\tan ^{-1}(0)-\tan ^{-1}(-\infty)\right]=\frac{1}{\sqrt{2}} \times \frac{\pi}{2} \\ & \Rightarrow A=2 B \end{aligned} $

We have, $I=\int_0^1 \sqrt{\frac{2+x}{2-x}} d x$

Let $\frac{2+x}{2-x}=u$

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

Now, lower limit $=1$

and upper limit $=3$

$ =\int_1^3 \frac{\sqrt{u}}{(u+1)^2} d u $

Now, let $s=\sqrt{u}$

$ \begin{aligned} d s & =\frac{1}{2 \sqrt{u}} \cdot d u=8 \int_1^{\sqrt{3}} \frac{s^2}{\left(s^2+1\right)^2} d s \\ & =8 \int_1^{\sqrt{3}}\left(\frac{1}{s^2+1}-\frac{1}{\left(s^2+1\right)^2}\right) d s \\ & =8\left|\tan ^{-1}(s)\right|_1^{\sqrt{3}}-8 \int_1^{\sqrt{3}} \frac{1}{\left(1+s^2\right)^2} d s \\ & =8\left(\tan ^{-1}(\sqrt{3})-\tan ^{-1}(1)\right)-8 \end{aligned} $

$ \int_1^{\sqrt{3}} \frac{1}{\left(1+s^2\right)^2} d s $

$ \begin{aligned} & =8\left(\frac{\pi}{3}-\frac{\pi}{4}\right)-8 \int_1^{\sqrt{3}} \frac{1}{\left(1+s^2\right)^2} d s \\ & =\frac{2 \pi}{3}-8 \int_{\pi / 4}^{\pi / 3} \cos ^2 p d p \end{aligned} $

$ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text { [put } s=\tan p] $

$ \begin{aligned} & =\frac{2 \pi}{3}+2-\sqrt{3}+1-\left.4 p\right|_{\pi / 4} ^{\pi / 3} \\ & =\frac{1}{3}(6-3 \sqrt{3}+\pi)=\frac{\pi}{3}+2-\sqrt{3} \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} & M=\int_0^{\infty} \frac{\log t}{1+t^3} \text { and } N=\int_{-\infty}^{\infty} \frac{t e^{2 t}}{1+e^{3 t}} d t \\ & N=\int_{-\infty}^{\infty} \frac{t \cdot e^{2 t}}{1+e^{3 t}} d t \end{aligned} \end{aligned} $

Let $e^t=x$

$e^t \cdot d t=d x$

Lower limit = 0

and Upper limit $=\infty$

$ \begin{aligned} N & =\int_0^{\infty} \frac{\log x \cdot x^2}{1+x^3} \cdot \frac{d x}{x}=\int_0^{\infty} \frac{x \log x}{1+x^3} d x \\ & =\int_0^{\infty} \frac{x \log x}{1+x^3} d x=-\int_0^{\infty} \frac{\log x}{1+x^3} d x \\ \Rightarrow N & =-\int_0^{\infty} \frac{\log t}{1+t^3} d t \Rightarrow N=-M \end{aligned} $

To solve the integral $ I = \int_{-2}^2 \left(4 - x^2\right)^{\frac{5}{2}} \, dx $, we can use a substitution method.

Substitution:

Substitute $ x = 2 \sin u $:

Then $ dx = 2 \cos u \, du $.

Change the limits:

When $ x = -2 $, $ u = -\frac{\pi}{2} $.

When $ x = 2 $, $ u = \frac{\pi}{2} $.

Substitute into the integral:

$ I = \int_{-\pi/2}^{\pi/2} \left(4 - (2\sin u)^2\right)^{\frac{5}{2}} \cdot 2\cos u \, du $

$ = \int_{-\pi/2}^{\pi/2} \left(4 - 4\sin^2 u\right)^{\frac{5}{2}} \cdot 2\cos u \, du $

Simplify:

Since $ 4 - 4\sin^2 u = 4\cos^2 u $, we have:

$ = \int_{-\pi/2}^{\pi/2} \left(4\cos^2 u\right)^{\frac{5}{2}} \cdot 2\cos u \, du $

$ = 32 \int_{-\pi/2}^{\pi/2} \cos^6 u \, du $

Using Reduction Formula:

Solving this integral using the reduction formula yields:

$ I = 64 \int_{-\pi/2}^{\pi/2} \cos^6 u \, du = 20\pi $

Thus, the value of the integral is $ I = 20\pi $.