Definite Integration

$ \begin{aligned} &\text { Given, } H(x)=3 x^4+6 x^3-2 x^2+1\\ &\begin{aligned} & \text { Given, } \frac{H(x)}{(x-1)(x-2)(x+1)} \\ & \begin{aligned} &=f(x)+\frac{g(x)}{(x-1)(x+1)(x-2} \text { taking LCM } \\ & H(x)=f(x)(x-1)(x+1)(x-2)+g(x) \\ & H(-1)=g(-1) \\ & H(2)=g(2) \\ & H(1)=g(1) \\ & \text { Now, } H(-1)+2 H(2)-3 H(1) \\ &=g(-1)+2 g(2-3 g(1) \end{aligned} \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { } \begin{aligned} I & =\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{\cos x-\sin x}{\sin 2 x} d x \\ I & =\int_{\sqrt{2}}^{\frac{\sqrt{3}}{2}+\frac{1}{2}} \frac{d t}{t^2-1} \end{aligned}\\ &\text { Put } \sin x+\cos x=t\\ &\begin{aligned} & (\cos x-\sin x) d x=d t \\ & 1+\sin 2 x=t^2 \end{aligned} \end{aligned} $

$ \begin{aligned} & =\left[\frac{1}{2} \ln \left(\frac{t-1}{t+1}\right)\right]_{\sqrt{2}}^{\frac{\sqrt{3}+1}{2}} \\ & =\frac{1}{2}\left[\ln \left(\frac{\frac{\sqrt{3}+1}{2}-1}{\frac{\sqrt{3}+1}{2}+1}\right)-\ln \left(\frac{\sqrt{2}-1}{\sqrt{2}+1}\right)\right] \\ & =\frac{1}{2}\left[\ln \frac{\sqrt{3}-1}{\sqrt{3}+3}-\ln \left(\frac{\sqrt{2}-1}{\sqrt{2}+1}\right)\right] \\ & =\frac{1}{2}\left[\ln \frac{2(2 \sqrt{3}-3)}{6}-\ln (3-2 \sqrt{2})\right] \\ & =\frac{1}{2}\left[\ln \frac{2 \sqrt{3}-3}{3}+\ln (3+2 \sqrt{2})\right] \\ & =\frac{1}{2}\left[\ln \left(\frac{2-\sqrt{3}}{\sqrt{3}}\right)+\ln (3+2 \sqrt{2})\right] \\ & =\frac{1}{2} \ln \left[\frac{(3+2 \sqrt{2})(2-\sqrt{3})}{\sqrt{3}}\right] \end{aligned} $

$ \text { } \begin{aligned} I & =\int_0^{\frac{\pi}{2}} \frac{\sin x}{1+\cos x+\sin x} d x \\ & =\int_0^{\frac{\pi}{2}} \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos ^2 \frac{x}{2}+2 \sin \frac{x}{2} \cos \frac{x}{2}} d x \\ & =\int_0^{\frac{\pi}{2}} \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}} d x \end{aligned} $

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

$I=\int_0^\pi \frac{x \sin x}{1+\cos ^2 x}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Apply property

$ \begin{aligned} & \int_a^b x f(x) d x=\frac{a+b}{2} \int f(x) d x \\ & \text { If } f(a+b-x)=f(x) \\ & \qquad I=\frac{\pi}{2} \int_0^\pi \frac{\sin x}{1+\cos ^2 x} d x \end{aligned} $

$ \begin{aligned} & \text { Put } \cos x=t \\ & \sin x d x=-d t \\ & \qquad \begin{aligned} I & =-\frac{\pi}{2} \int_{+1}^{-1} \frac{d t}{1+t^2} \\ & =\frac{\pi}{2} \int_{-1}^1 \frac{d t}{1+t^2}=\frac{\pi}{2} \cdot 2 \int_0^1 \frac{d t}{1+t^2} \\ & =\pi\left(\tan ^{-1} t\right)_0^1 \\ & =\pi \tan ^{-1} 1=\pi \times \frac{\pi}{4}=\frac{\pi^2}{4} \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { } \int_a^b e^x d x=\lim _{h \rightarrow 0} h\left[e^a+e^{a+h}+e^{a+2 h}+\ldots+e^{a+(n-1) h}\right] \\ & =\lim _{h \rightarrow 0} h e^a\left[1+e^h+e^{2 h}+e^{3 h}+\ldots+e^{(n-1) h}\right] \\ & =\lim _{h \rightarrow 0} h e^a\left\{\frac{\left(e^h\right)^n-1}{e^h-1}\right\} \\ & \quad\left[\because a+a r+\ldots+a r^{n-1}=a\left(\frac{r^n-1}{r-1}\right)\right] \end{aligned} $

$ \begin{aligned} & =\lim _{h \rightarrow 0} h e^a\left(\frac{e^{n h}-1}{e^h-1}\right) \\ & =\lim _{h \rightarrow 0} e^a\left\{\frac{e^{b-a}-1}{\left(\frac{e^h-1}{h}\right)}\right\}=e^a\left(e^{b-a}-1\right)=e^b-e^a \end{aligned} $

We have, $I=\int_0^{\frac{\pi}{2}} \log (\sin x) d x$

Using the property:

$ \int_0^a f(x) d x=\int_0^a f(a-x) d x $

we replace $x$ with $\frac{\pi}{2}-x$, we get

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

$\begin{aligned} & \int_\limits{\frac{1}{4}}^{\frac{3}{4}} \cos \left(2 \cot ^{-1} \sqrt{\frac{1-x}{1+x}}\right) d x \\ & x=\cos 2 \theta \\ & \Rightarrow d x=(-2 \sin 2 \theta \mathrm{d} \theta) \end{aligned}$

Take limit as $\alpha$ and $\beta$

$\begin{aligned} & -2 \int_\limits\alpha^\beta \cos 2 \theta \cdot \sin 2 \theta d \theta \\ & =\int_\limits\alpha^\beta \sin 4 \theta d \theta \\ & =\left.\frac{-\cos 4 \theta}{4}\right|_\alpha ^\beta \\ & =-\left.\frac{1}{4}\left(2 \cdot\left(x^2\right)-1\right)\right|_{1 / 4} ^{3 / 4} \\ & =-\left.\frac{1}{4}\left(2 x^2-1\right)\right|_{1 / 4} ^{3 / 4} \\ & =-\frac{1}{4}\left(\frac{18}{16}-1-\frac{2}{16}+1\right) \\ & =-\frac{1}{4} \end{aligned}$

$\lim _\limits{x \rightarrow \frac{\pi}{2}}\left(\frac{\int_{x^3}^{(\pi / 2)^3}\left(\sin \left(2 t^{1 / 3}\right)+\cos \left(t^{1 / 3}\right)\right) d t}{\left(x-\frac{\pi}{2}\right)^2}\right)$

Using Newton Leibniz theorem

$\begin{aligned} & =\lim _\limits{x \rightarrow \frac{\pi}{2}}\left[\frac{\sin \left(2 \times \frac{\pi}{2}\right) \cdot 0-\sin (2 x) \cdot 3 x^2+\left(\cos \frac{\pi}{2}\right) \cdot 0-\cos x \cdot 3 x^2}{2\left(x-\frac{\pi}{2}\right)}\right] \\ & =\lim _\limits w{x \rightarrow \frac{\pi}{2}} \frac{-3 x^2 \sin 2 x-3 x^2 \cos x}{2\left(x-\frac{\pi}{2}\right)}\left(\frac{0}{0}\right) \text { form } \\ & =\lim _\limits{x \rightarrow \frac{\pi}{2}}\left[\frac{-6 x \sin 2 x-6 x^2 \cos 2 x-6 x \cos x+3 x^2 \sin x}{2}\right] \\ & =\frac{6 \times \frac{\pi^2}{4}+3 \times \frac{\pi^2}{4}}{2} \\ & =\frac{9 \pi^2}{8} \end{aligned}$

$\begin{aligned} & \int_\limits{-1}^2 \log _e\left(x+\sqrt{x^2+1}\right) d x \\ & =\left[x \log _e\left(x+\sqrt{x^2+1}\right)\right]_{-1}^2-\int_\limits{-1}^2 \frac{x}{\left(x+\sqrt{x^2+1}\right)}\left(1+\frac{x}{\sqrt{x^2+1}}\right) d x \\ & =2 \log _2(2+\sqrt{5})+\log _e(\sqrt{2}-1)-\int_\limits{-1}^2 \frac{x}{\sqrt{x^2+1}} d x \\ & =\log _e\left[(2+\sqrt{5})^2(\sqrt{2}-1)\right]-\left[\sqrt{x^2+1}\right]_{-1}^2 \\ & =\log _e\left[\frac{9+4 \sqrt{5}}{1+\sqrt{2}}\right]-\sqrt{5}+\sqrt{2} \\ & =\sqrt{2}-\sqrt{5}+\log _e\left[\frac{9+4 \sqrt{5}}{1+\sqrt{2}}\right] \end{aligned}$

$\begin{aligned} & \text { Let } \sqrt{e^x-1}=t \\ & e^x-1=t^2 \\ & e^x=1+t^2 \\ & e^x=0+2 t-\frac{d t}{d x} \\ & \frac{d t}{d x}=\frac{e^x}{2 t}=\frac{t^2+1}{2 t} \\ & I=\int \frac{2 t}{t\left(1+t^2\right)} d t=2 \tan ^{-1} t \\ & \Rightarrow \quad I=\int_\limits\alpha^{\log ^4} \frac{d x}{\sqrt{e^x-1}} \\ & I=\left.2 \tan ^{-1} \sqrt{e^x-1}\right|_\alpha ^{\log 4} \\ & =2\left(\tan ^{-1} \sqrt{3}-\tan ^{-1} \sqrt{e^\alpha-1}\right)=\frac{\pi}{6} \\ & \Rightarrow \frac{\pi}{3}-\tan ^{-1}\left(\sqrt{e^\alpha-1}\right)=\frac{\pi}{12} \\ & \tan ^{-1}\left(\sqrt{e^\alpha-1}\right)=\frac{\pi}{4} \\ & \Rightarrow e^\alpha-1=1 \\ & e^\alpha=2 \Rightarrow e^{-\alpha}=\frac{1}{2} \end{aligned}$

$\therefore \quad$ Quadratic equation whose roots are $e^a$ & $e^{-\alpha}$ is

$\begin{aligned} & x^2-\left(e^\alpha+e^{-\alpha}\right) x+e^\alpha \times e^{-\alpha}=0 \\ & x^2-\left(2+\frac{1}{2}\right) x+1=0 \\ & 2 x^2-5 x+2=0 \end{aligned}$

$\begin{aligned} & I(21)=\int_\limits0^1\left(1-x^k\right)^{21} d x \\ & =\int_\limits0^1\left(1-x^k\right)\left(1-x^k\right)^{20} d x \\ & =\int_\limits0^1\left(1-x^k\right)^{20} d x-\int_0 x^k\left(1-x^k\right)^{20} d x \\ & I(21)=I(20)-\int_\limits0^1 x^k\left(1-x^k\right)^{20} d x \end{aligned}$

$I(21)=I(20)-\left\lfloor\frac{\left(1-x^k\right)^{21}}{-21 k} x-\int_\limits0^1 \frac{(1-x^k)^{21}}{-21 k} d x\right\rfloor$

$\begin{aligned} & I(21)=I(20)-\frac{1}{21 k} I(20) \\ & \Rightarrow[I(21)](21 k+1)=21 K I(20) \\ & \Rightarrow 21 K=147 \Rightarrow K=7 \end{aligned}$

$\begin{aligned} & \int_\limits0^{\pi / 4} \frac{\cos ^2 x \cdot \sin ^2 x}{\left(\cos ^3 x+\sin ^3 x\right)^2} d x \\ & =\int_\limits0^{\pi / 4} \frac{\tan ^2 x \cdot \sec ^2 x}{\left(1+\tan ^3 x\right)^2} d x \end{aligned}$

Let $\tan x=t$

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

Let $1+t^3=\mathrm{z}$

$\begin{gathered} 3 t^2 d t=d z \\ \frac{1}{3} \int_\limits1^2 \frac{d z}{z^2}=\left.\frac{1}{3}\left(-\frac{1}{z}\right)\right|_1 ^2 \\ =-\frac{1}{3}\left(\frac{1}{2}-1\right)=\frac{1}{6}\end{gathered}$

First, let's rewrite the given integral using the given form of the Beta function. The given integral is:

$\int_\limits0^1\left(1-x^{10}\right)^{20} \mathrm{~d} x$

To use the Beta function, let us make a substitution. Let $ x^{10} = t $. Then, $ dx = \frac{1}{10}t^{-\frac{9}{10}} dt $ or $ dx = \frac{1}{10} t^{-\frac{9}{10}} dt $. The limits of integration change as follows: when $ x = 0 $, $ t = 0 $, and when $ x = 1 $, $ t = 1 $.

Substituting these into the integral, we have:

$\int_\limits0^1 (1 - t)^{20} \cdot \frac{1}{10} t^{-\frac{9}{10}} dt$

which simplifies to:

$\frac{1}{10} \int_\limits0^1 (1 - t)^{20} t^{-\frac{9}{10}} dt$

We recognize this integral as a Beta function $ \beta(m, n) $ where $ m = 1 - \frac{9}{10} = \frac{1}{10} $ and $ n = 20 + 1 = 21 $.

Therefore, we can write this as:

$\frac{1}{10} \beta \left( \frac{1}{10}, 21 \right)$

Comparing this to $ a \times \beta(b, c) $, we have $ a = \frac{1}{10} $, $ b = \frac{1}{10} $, and $ c = 21 $.

Now we calculate $ 100(a + b + c) $:

$100 \left( \frac{1}{10} + \frac{1}{10} + 21 \right) = 100 \left( \frac{1}{10} + \frac{1}{10} + 21 \right) = 100 \left( \frac{1}{5} + 21 \right) = 100 \left( \frac{1}{5} + \frac{105}{5} \right) = 100 \left( \frac{106}{5} \right) = 100 \times 21.2 = 2120$

So, the answer is Option D, 2120.

$\int_0^{\pi / 4} \frac{136 \sin x}{3 \sin x+5 \cos x} d x$

$\begin{aligned} & \sin x=A(3 \sin x+5 \cos x)+B(3 \cos x-5 \sin x) \\ & \begin{array}{l} 3 A-5 B=1 \\ 5 A+3 B=0 \end{array}>A=\frac{3}{34} \quad B=\frac{-5}{34} \end{aligned}$

$\begin{aligned} & \int_0^{\pi / 4} \frac{136\left[\frac{3}{34}(3 \sin x+5 \cos x)-\frac{5}{34}(3 \cos x-5 \sin x)\right]}{3 \sin x+5 \cos x} d x \\ & \int_0^{\pi / 4} 12 d x-20 \int_0^{\pi / 4} \frac{3 \cos x-5 \sin x}{3 \sin x+5 \cos x} d x \\ & 12 \times \frac{\pi}{4}-20\left[\ln \left|\frac{3}{\sqrt{2}}+\frac{5}{\sqrt{2}}\right|-\ln 5\right] \\ & 3 \pi-20 \ln 2^{5 / 2}+20 \ln 5 \\ & \Rightarrow 3 \pi-50 \ln 2+20 \ln 5 \end{aligned}$

$\begin{aligned} & I=\int_\limits{-\pi}^\pi \frac{2 y(1+\sin y)}{1+\cos ^2 y} d y \\ & =\int_\limits0^\pi\left(\frac{2 y(1+\sin y)}{1+\cos ^2 y}+\frac{-2 y(1-\sin y)}{1+\cos ^2 y}\right) d y \\ & =\int_\limits0^\pi\left(\frac{2 y+2 y \sin y-2 y+2 y \sin y}{1+\cos ^2 y}\right) d y \end{aligned}$

$I=4 \int_0^\pi\left(\frac{y \sin }{1+\cos ^2 y}\right) d y \quad \text{... (1)}$

$\begin{aligned} & I=4 \int_\limits0^\pi\left(\frac{(\pi-y) \sin (\pi-)}{1+\cos ^2(\pi-y)}\right) d y \\ & I=4\left\lfloor\int_\limits0\left(\frac{\pi \sin y}{1+\cos ^2 y}\right) d y-\int_\limits0^\pi \frac{y \sin y}{1+\cos ^2 y} d y\right\rfloor \quad \text{... (2)} \end{aligned}$

Adding equation (1) and (2)

$\begin{aligned} & 2 I=4 \int_\limits0^\pi\left(\frac{\pi \sin y}{1+\cos ^2 y}\right) d y \\ & I=2 \pi \int_\limits0^\pi \frac{\sin y}{1+\cos ^2 y} d y \\ & =2 \pi \times \frac{\pi}{2} \\ & =\pi^2 \end{aligned}$

Given $f(x)=\int_\limits0^x\left(t+\sin \left(1-e^t\right)\right) d t$

Now, $\lim _\limits{x \rightarrow 0} \frac{f(x)}{x^3}\left(\frac{0}{0} \text { form }\right)$

$\begin{aligned} & =\lim _{x \rightarrow 0} \frac{\int_\limits0^x\left(t+\sin \left(1-e^t\right)\right) d t}{x^3} \\ & =\lim _{x \rightarrow 0} \frac{x+\sin \left(1-e^x\right)}{3 x^2}\left(\frac{0}{0}\right) \\ & =\lim _{x \rightarrow 0} \frac{1+\cos \left(1-e^x\right)\left(-e^x\right)}{6 x}\left(\frac{0}{0}\right) \\ & =\lim _{x \rightarrow 0} \frac{-\sin \left(1-e^x\right)\left(e^x\right)^2+\cos \left(1-e^x\right)\left(-e^x\right)}{6} \\ & =-\frac{1}{6} \end{aligned}$

$\begin{aligned} & \text { Given, } \int_\limits{-1}^1 \frac{\cos \alpha x}{1+3^x} d x=\frac{2}{\pi} \\ & \begin{aligned} I & =\int_\limits{-1}^1 \frac{\cos \alpha x}{1+3^x} d x \\ \Rightarrow I & =\int_\limits0^1\left(\frac{\cos \alpha x}{1+3^x}+\frac{\cos \alpha x}{1+3^{-x}}\right) d x \\ & =\int_\limits0^1 \cos \alpha x d x \\ & =\left(\frac{\sin \alpha x}{\alpha}\right)_0^1 \\ & =\frac{\sin \alpha}{\alpha} \\ \Rightarrow & \frac{\sin \alpha}{\alpha}=\frac{2}{\pi} \\ \Rightarrow & \alpha=\frac{\pi}{2} \end{aligned} \end{aligned}$

$f(x)=\left\{\begin{array}{cc} -2 & -2 \leq x \leq 0 \\ x-2 & 0< x \leq 2 \end{array} h(x)=f|x|+|f(x)|\right.$

$\begin{aligned} & h(x)=\left\{\begin{array}{cc} -x-2+2=-x & -2 \leq x \leq 0 \\ 0 & 0< x \leq 2 \end{array}\right. \\ & \therefore \int_\limits{-2}^2 h(x) d x=\int_\limits{-2}^0-x d x+\int_\limits0^2 0 d x \\ & \left.\frac{x^2}{2}\right|_{-2} ^0=\frac{4}{2}=2 \end{aligned}$

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)

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