Definite Integration

$ \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} $

$ \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} $

$ \text { } \begin{aligned} \int_0^{\pi / 2} \frac{x \tan x \sec ^2 x}{\tan ^4 x+1} d x \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\ =\int_0^{\pi / 2} \frac{x \sin x}{\frac{\cos ^3 x}{\sin ^4 x+\cos ^4 x} x} d x \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\ =\int_0^{\pi / 2} \frac{x \sin x \cos x}{\sin ^4 x+\cos ^4 x} d x \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ =\int_0^{\pi / 2} \frac{\left(\frac{\pi}{2}-x\right) \sin \left(\frac{\pi}{2}-x\right) \cos \left(\frac{\pi}{2}-x\right)}{\sin ^4\left(\frac{\pi}{2}-x\right)+\cos ^4\left(\frac{\pi}{2}-x\right)} d x \\ =\int_0^{\pi / 2} \frac{\left(\frac{\pi}{2}-x\right) \cos x \sin x}{\sin ^4 x+\cos ^4 x} d x \end{aligned} $

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

$ 2 I=\frac{\pi}{2} \int_0^{\pi / 2} \frac{\sin x \cos x}{\sin ^4 x+\cos ^4 x} d x $

Put, $\sin ^2 x=t$

$ \Rightarrow 2 \sin x \cos x=d t $

When, $x=0, t=0$

and $x=\frac{\pi}{2}, t=1$

$ \begin{aligned} I & =\frac{\pi}{4} \int_0^1 \frac{\frac{1}{2} d t}{2 t^2-2 t+1} \\ \Rightarrow I & =\frac{\pi}{8} \int_0^1 \frac{d t}{\left(t-\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^2} \\ & =\frac{\pi}{8}\left[\tan ^{-1}\left(\frac{t-\frac{1}{2}}{\frac{1}{2}}\right)\right]_0^1 \\ & =\frac{\pi}{8}\left[\tan ^{-1}(1)-\tan ^{-1}(-1)\right] \\ & =\frac{\pi}{8}\left(\frac{\pi}{4}+\frac{\pi}{4}\right)=\frac{\pi^2}{16} \end{aligned} $

Let $I=\int_3^6 \frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}} d x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Using properties

$ \begin{aligned} \int_a^b f(x) d x & =\int_a^b f(a+b-x) d x \\ I & =\int_3^6 \frac{\sqrt{9-x}}{\sqrt{x}+\sqrt{9-x}} d x \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

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

$ 2 I=\int_3^6 d x \Rightarrow I=\frac{1}{2}[x]_3^6=\frac{3}{2} $

$\ln P=$

$ \begin{aligned} & \frac{1}{n}\left[\begin{array}{r} \ln \left(1+\frac{1}{n^2}\right)+\ln \left(1+\frac{2^2}{n^2}\right)+\ln \left(1+\frac{3^2}{n^2}\right) \\ \ldots \ln \left(1+\frac{n^2}{n^2}\right) \end{array}\right] \\ & \Rightarrow \ln P=\sum_{r=1}^n \ln \left(1+\frac{r^2}{n^2}\right) \frac{1}{n} \\ & \Rightarrow \ln P=\int_0^1 \ln \left(1+x^2\right) d x\left\{\begin{array}{l} r / n=x \\ 1 / n=d x \end{array}\right. \end{aligned} $

Using by parts,

$ \begin{aligned} & \ln P=\left[\ln \left(1+x^2\right) \cdot x-\int \frac{1}{1+x^2} \cdot 2 x \cdot x d x\right]_0^1 \\ & \Rightarrow \ln P=\ln 2-0-2 \int_0^1 \frac{1+x^2-1}{1+x^2} d x \\ & =\ln 2-2\left[x-\tan ^{-1} x\right]_0^1=\ln 2-2+\frac{\pi}{2} \\ & \Rightarrow P=e^{\left(\ln 2+\frac{\pi}{2}-2\right)}=2 e^{\frac{\pi-4}{2}} \end{aligned} $

$ \begin{aligned} & \text { Let } I=\int_{-1}^1 x|x| d x \\ & |x|=\left\{\begin{array}{cc} -x & ; x<0 \\ x & ; x>0 \end{array}\right. \\ & I=\int_{-1}^0-x^2 d x+\int_0^1 x^2 d x=\left[\frac{-x^3}{3}\right]_{-1}^0+\left[\frac{x^3}{3}\right]_0^1 \\ & =0-\left[\frac{-(-1)^3}{3}\right]+\left[\frac{1}{3}\right] \\ & I=\frac{-1}{3}+\frac{1}{3}=0 \end{aligned} $

Let

$ \begin{aligned} I & =\int_{-\pi / 2}^{\pi / 2} \sin ^2 x \cos ^2 x(\sin x+\cos x) d x \\ & =\int_{-\pi / 2}^{\pi / 2} \sin ^3 x \cos ^2 x d x+\int_{-\pi / 2}^{\pi / 2} \sin ^2 x \cos ^3 x d x \end{aligned} $

Now, we know that for an odd function

$ f(x)=\int_{-a}^a f(x) d x=0 $

And for an even function $g(x)$,

$ \int_{-a}^a g(x) d x=2 \int_0^a g(x) d x $

Now, $f(x)=\sin ^3 x \cdot \cos ^2 x$ is an odd function and $g(x)=\sin ^2 x \cdot \cos ^3 x$ is an even function.

Hence, first integration is 0 .

For second part, we get

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

Now, let $t=\sin x$

$ d t=\cos x d x $

And for $x=0, t=0$ and for $x=\frac{\pi}{2}, t=1$

Hence, integration becomes,

$ \begin{aligned} 2 \int_0^1 t^2\left(1-t^2\right) & d t=2 \int_0^1\left(t^2-t^4\right) d t \\ = & 2\left[\frac{t^3}{3}-\frac{t^5}{5}\right]_0^1=2\left[\frac{1}{3}-\frac{1}{5}\right]=\frac{4}{15} \end{aligned} $

$ \begin{aligned} &\\ &\text { } \begin{aligned} & \int_0^3\left(3 x^2-4 x+2\right) d x=k \\ & \quad\left\{\because \int x^n d x=\frac{x^{n+1}}{n+1}+C\right\} \end{aligned} \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad 3 \int_0^3 x^2 d x-4 \int_0^3 x d x+2 \int_0^3 d x=k \\ & \Rightarrow \quad 3\left[\frac{x^3}{3}\right]_0^3-4\left[\frac{x^2}{2}\right]_0^3+2[x]_0^3=k \\ & \Rightarrow \quad 3\left[\frac{3^3}{3}-0\right]-4\left[\frac{3^2}{2}-0\right]+2[3-0]=k \\ & \Rightarrow \quad 27-18+6=k \Rightarrow k=15 \\ & \Rightarrow \quad 3 x^2-4 x+2=3 k / 5 \\ & \Rightarrow \quad 3 x^2-4 x+2=\frac{3 \times 15}{5} \\ & \Rightarrow \quad 3 x^2-4 x+2=9 \\ & \Rightarrow \quad 3 x^2-4 x-7=0 \\ & \Rightarrow \quad 3 x^2-7 x+3 x-7=0 \\ & \Rightarrow \quad(3 x-7)(x+1)=0 \\ & \Rightarrow \quad x=\frac{7}{3}, x=-1 \end{aligned} $

$ \begin{aligned} & \text { Let } I=\int_0^\pi \frac{x \cos ^2 x}{1+\sin x} d x \\ & =\int_0^\pi \frac{x\left(1-\sin ^2 x\right)}{1+\sin x} d x \\ & =\int_0^\pi \frac{x(1-\sin x)(1+\sin x)}{1+\sin x} d x \\ & =\int_0^\pi x(1-\sin x) d x \\ & =\int_0^\pi x d x+\int_0^\pi x(-\sin x) d x \\ & =\left|\frac{x^2}{2}\right|_0^\pi+\left|x \cos x-\int \cos x d x\right|_0^\pi \\ & =\left|\frac{x^2}{2}\right|_0^\pi+|x \cos x-\sin x d x|_0^\pi \\ & =\frac{\pi^2}{2}+\pi \cos \pi-\sin \pi-0+\sin 0 \\ & =\frac{\pi^2}{2}-\pi-0+0=\frac{\pi^2-2 \pi}{2}=\frac{\pi(\pi-2)}{2} \end{aligned} $

$ \begin{aligned} & \text { Let } I=\int_{-2}^2[2-x] d x \\ & =\int_{-2}^{-1} 3 d x+\int_{-1}^0 2 d x+\int_0^1 1 d x+\int_1^2 0 d x \end{aligned} $

$ \begin{aligned} & =|3 x|_{-2}^1+|2 x|_{-1}^0+|x|_0^1+0 \\ & =3(-1+2)+2(0+1)+(1-0) \\ & =3 \times 1+2 \times 1+1=3+2+1=6 \end{aligned} $

Let $I=\int_0^2 \frac{x}{(2-x)^{3 / 4}} d x$

Let $2-x=v$

$ \Rightarrow-d x=d v \Rightarrow d x=-d v $

If $\quad x=0 \Rightarrow v=2$

and if $x=2 \Rightarrow v=0$

$ \begin{aligned} & \therefore \quad I=-\int_2^0 \frac{2-v}{v^{3 / 4}} d v \\ & =\int_0^2\left[\frac{2}{v^{3 / 4}}-\frac{v}{v^{3 / 4}}\right] d v \\ & =2 \int_0^2 v^{-3 / 4} d v-\int_0^2 v^{1 / 4} d v \\ & =2\left|\frac{v^{1 / 4}}{1 / 4}\right|_0^2-\left|\frac{v^{5 / 4}}{5 / 4}\right|_0^2=8(2)^{1 / 4}-\frac{4}{5}(2)^{5 / 4} \\ & =2^{1 / 4}\left[8-\frac{4 \times 2}{5}\right]=2^{1 / 4}\left[\frac{40-8}{5}\right]=(32 / 5) 2^{1 / 4} \end{aligned} $

Given, $\int_0^2 x^3(2-x)^4 d x$

Let $2-x=z \Rightarrow-d x=d z \Rightarrow d x=-d z$

If $\quad 2-0=z$

$ \Rightarrow \quad z=2 \text { and } 2-2=z \Rightarrow z=0 $

$ \therefore-\int_2^0(2-z)^3 z^4 d z=\int_0^2(2-z)^3 z^4 d z $

$ =\int_0^2\left(8-z^3-12 z+6 z^2\right) z^4 d z $

$ =\int_0^2\left(8 z^4-z^7-12 z^5+6 z^6\right) d z $

$ =\left|\frac{8 z^5}{5}-\frac{z^8}{8}-\frac{12 z^6}{6}+\frac{6 z^7}{7}\right|_0^2 $

$ =\left|\frac{8}{5} z^5-\frac{z^8}{8}-2 z^6+\frac{6}{7} z^7\right|_0^2 $

$ =\frac{8}{5}(2)^5-\frac{(2)^8}{8}-2(2)^6+\frac{6}{7}(2)^7 $

$ =\frac{256}{5}-32-128+\frac{6}{7} \times 128 $

$ =\frac{7 \times 256-160 \times 5 \times 7+5 \times 6 \times 128}{35}=\frac{32}{35} $

$ \begin{aligned} & \text {Let } I=\int_0^3\left|x^2-3 x+2\right| d x \\ & \begin{array}{r} x^2-3 x+2=\left\{\begin{array}{cc} x^2-3 x+2, & 0 < x < 1 \\ -\left(x^2-3 x+2\right), & 1 < x < 2 \\ \left(x^2-3 x+2\right), & 2 < x < 3 \end{array}\right. \\ =\int_0^1\left(x^2-3 x+2\right) d x-\int_1^2\left(x^2-3 x+2\right) \\ d x+\int_2^3\left(x^2-3 x+2\right) d x \end{array} \\ & =\left(\frac{x^3}{3}-\frac{3 x^2}{2}+2 x\right)_0^1-\left(\frac{x^3}{3}-\frac{3 x^2}{2}+2 x\right)_1^2 +\left(\frac{x^3}{3}-\frac{3 x^2}{2}+2 x\right)_2^3 \end{aligned} $

$ \begin{aligned} & =\left(\frac{1}{3}-\frac{3}{2}+2\right)-\left(\frac{8}{3}-\frac{12}{2}+4\right)+ \left(\frac{1}{3}-\frac{3}{2}+2\right)+\left(\frac{27}{3}-\frac{27}{2}+6\right)-\left(\frac{8}{3}-\frac{12}{2}+4\right) \\ & =11 / 6 \end{aligned} $

Let

$ \begin{aligned} & I=\int_{\frac{-\pi}{8092}}^{\frac{\pi}{8092}} \frac{\sec (2023) x}{1+(2023)^{(2023 x)}} d x \\ & \int_a^b f(x) d x=\int_a^b f(a+b-x) d x \\ & I=\int_{\frac{\pi}{8092}}^{\frac{\pi}{8092}} \frac{(2023)^{(2023 x)} \sec (2023) x}{1+(2023)^{2023 x}} d x \ldots \end{aligned} $

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

$ \begin{aligned} &2I=\int_{-\frac{\pi}{8092}}^{\frac{\pi}{8092}} \sec (2023 x) d x \\ & I=\int_0^{\frac{\pi}{8092}} \sec (2023 x) d x \\ = & {\left[\frac{\log / \sec (2023 x)+\tan (2023 x)}{2023}\right]_0^{\frac{\pi}{8092}}+C } \\ = & \frac{\log (\sqrt{2}+1)}{2023}+C \end{aligned} $

$ \text { } \begin{aligned} I & =\int_0^2 x^{5 / 2} \sqrt{2-x} d x \\ \text { Let } x & =2 \sin ^2 \theta \\ d x & =4 \sin \theta \cos \theta d \theta \end{aligned} $

$ \begin{aligned} & \text { At } x=0, \theta=0 \\ & \text { At } x=2, \theta=\pi / 2 \\ & \begin{aligned} \therefore \quad & =\int_0^{\pi / 2}\left(2 \sin ^2 \theta\right)^{5 / 2} \sqrt{2-2 \sin ^2 \theta} \\ & 4 \sin \theta \cos \theta d \theta \\ = & \int_0^{\pi / 2} 2^{5 / 2} \sin ^5 \theta \cdot 2^{1 / 2} \cos \theta 4 \sin \theta \cos \theta d \theta \\ = & 32 \int_0^{\pi / 2} \sin ^6 \theta \cos ^2 \theta d \theta \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { Using Walli's formula, }\\ &\begin{aligned} \int_0^{\pi / 2} & \sin ^m x \cos ^n x d x \\ & =\frac{[(m-1)(m-3)(m-5) \ldots \ldots][(n-1)(n-3) \ldots .]}{[(m+n)(m+n-2)(m+n-4) \ldots . . k]} \cdot k \\ k & =\left\{\begin{array}{l} \frac{\pi}{2}, m \text { and } n \text { are even } \\ 1, \text { otherwise } \end{array}\right. \\ & =32 \cdot \frac{(5 \cdot 3 \cdot 1)(1)}{8 \cdot 6 \cdot 4 \cdot 2} \times \frac{\pi}{2}=\frac{5 \pi}{8} \end{aligned} \end{aligned} $

$ \text { } \begin{aligned} & \int_0^{\frac{\pi}{4}} \frac{\sec x d x}{1+2 \sin ^2 x} \\ & =\int_0^{\pi / 4} \frac{\cos x d x}{\cos ^2 x\left(1+2 \sin ^2 x\right)} \end{aligned} $

$ \begin{aligned} &=\int_0^{\pi / 4} \frac{\cos x d x}{\left(1-\sin ^2 x\right)\left(1+2 \sin ^2 x\right)}\\ &\text { Let } \sin x=t \Rightarrow \cos x d x=d t\\ &\begin{aligned} & =\int_0^{1 / \sqrt{2}} \frac{d t}{\left(1-t^2\right)\left(1+2 t^2\right)} \\ & =\int_0^{1 / \sqrt{2}}\left(\frac{A}{1-t}+\frac{B}{1+t}+\frac{C t+D}{1+2 t^2}\right) d t \end{aligned} \end{aligned} $

$ \begin{aligned} & \begin{array}{r} =\int_0^{\frac{1}{\sqrt{2}}} \frac{1}{6}\left(\frac{1}{1-t}\right) d t+\frac{1}{6} \int_0^{\frac{1}{\sqrt{2}}} \frac{d t}{t+1}+\frac{2}{3} \int_0^{\frac{1}{\sqrt{2}}} \frac{d t}{1+2 t^2} \\ \left(\because A=\frac{1}{6}=B \text { and } C=0, D=\frac{2}{3}\right) \end{array} \\ & \Rightarrow-\frac{1}{6}[\ln (1-t)]_0^{\frac{1}{\sqrt{2}}}+\frac{1}{6}[\ln (1+t)]_0^{\frac{1}{\sqrt{2}}} \\ & \quad+\frac{2}{3} \cdot \frac{1}{2} \cdot \frac{1}{\sqrt{2}}\left[\tan ^{-1}(\sqrt{2} t)\right] \\ & =-\frac{1}{6} \ln \left(1-\frac{1}{\sqrt{2}}\right)+\frac{1}{6}\left(\ln \left(1+\frac{1}{\sqrt{2}}\right)\right)+\frac{1}{3 \sqrt{2}} \tan ^{-1}(1) \\ & =\frac{1}{3} \ln (\sqrt{2}+1)+\frac{1}{3 \sqrt{2}} \cdot \frac{\pi}{4} \\ & =\frac{1}{3} \ln (\sqrt{2}+1)+\frac{\pi \sqrt{2}}{12} \end{aligned} $

$ \begin{aligned} &\text { Now, }\\ &\begin{aligned} & \lim _{n \rightarrow \infty}\left[\frac{1}{n^2} \sec ^2 \frac{1}{n^2}+\frac{2}{n^2} \sec ^2 \frac{4}{n^2}+\ldots+\frac{1}{n} \sec ^2 1\right] \\ & =\lim _{n \rightarrow \infty} \frac{1}{n} \cdot \frac{n}{n} \cdot \sec ^2 \frac{n^2}{n^2}=\int_0^1 x \sec ^2 x^2 d x \\ & =\frac{1}{2} \int_0^1 2 x \sec ^2 x^2 d x \\ & \text { Let } x^2=z \Rightarrow 2 x d x=d z \\ & \qquad=\frac{1}{2} \int_0^1 \sec ^2 z d z=\frac{1}{2}(\tan z)_0^1 \\ & =\frac{1}{2} \tan 1 \end{aligned} \end{aligned} $

Given, $\int_2^5 \sqrt{\frac{5-x}{x-2}} d x$

Let $x$

$ \begin{aligned} x-2 & =z^2 \Rightarrow d x=2 z d z \\ & =\int_0^{\sqrt{3}} \frac{\sqrt{5-\left(z^2+2\right.}}{z} 2 z d z \\ & =2 \int_0^{\sqrt{3}} \sqrt{3-z^2} d z \\ & =2\left[\frac{z \sqrt{3-z^2}}{2}=\frac{3}{2} \sin ^{-1} \frac{z}{\sqrt{3}}\right]_0^{\sqrt{3}} \\ & =2\left[\frac{3}{2} \sin ^{-1}(1)\right]-3 \sin ^{-1}(1)=\frac{3 \pi}{2} \end{aligned} $

Given, $\int_0^{\frac{\pi}{2}} \sin ^6 x \cos ^4 x d x$

Using Wallis formula,

$ \begin{aligned} & \int_0^{\frac{\pi}{2}} \sin ^m x \cos ^n x d x, m, n \in z \\ & (m-1)(m-3) \ldots 2 \text { or } \cdot(n-1) \\ & =\frac{(n-3) \ldots 2 \text { or } 1}{(m+n)(m+n-2) \ldots 2 \text { or } 1} \times \lambda \end{aligned} $

where, $\lambda=\frac{\pi}{2}, m$ and $n$ both even.

$ \text { So, Let } \begin{aligned} \mathrm{I} & =\int_0^{\frac{\pi}{2}} \sin ^6 x \cos ^4 x d x \\ & =\frac{5 \times 3 \times 1 \times 3 \times 1}{10 \times 8 \times 6 \times 4 \times 2} \times \frac{\pi}{2} \\ & =\frac{3 \pi}{512} \end{aligned} $

$ \begin{aligned} & \ \int_{1 / 2}^2\left|\log _{10} x\right| d x=\int_{1 / 2}^1\left(-\log _{10} x\right) d x \\ & +\int_1^2 \log _{10} x d x \\ & =-\int_{1 / 2}^1\left(\log _e x \cdot \log _{10} e\right) d x \\ & +\int_1^2\left(\log _e x \cdot \log _{10} e\right) d x \\ & =-\log _{10} e \int_{1 / 2}^1(\log x) d x+\log _{10} e \int_1^2(\log x) d x \\ & =-\log _{10} e[x(\log x-1)]_{\frac{1}{2}}^1 \\ & +\log _{10} e[x(\log x-1)]_1^2 \quad \text { (using by parts) } \\ & =-\log _{10} e\left[1(0-1)-\frac{1}{2}\left(\log \frac{1}{2}-1\right)\right] \\ & +\log _{10} e[2(\log 2-1)-1(0-1)] \\ & =\log _{10} e\left[\left(1+\frac{1}{2} \log \frac{1}{2}-\frac{1}{2}\right)+(2 \log 2-1)\right] \\ & =\left(\log _{10} e\right)\left(-\frac{1}{2} \log 2+2 \log 2-\frac{1}{2}\right) \\ & =\log _{10} e\left(\frac{3}{2} \log 2-\frac{1}{2}\right) \\ & =\left(\log _{10} e\right)\left(\frac{3}{2} \log _e 2-\frac{1}{2} \log _e e\right) \\ & =\frac{1}{2} \log _{10} e\left(\log _e 8-\log _e e\right) \\ & =\frac{1}{2}\left(\log _{10} e\right)\left(\log _e\left(\frac{8}{e}\right)\right)=\frac{1}{2} \log _{10}\left(\frac{8}{e}\right) \end{aligned} $

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

Adding Eqs. (i) and (ii), we get

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

$ I=\int_0^{2 \pi}[|\sin x|+|\cos x|] d x$

$ \begin{aligned} & \text { Period of }|\sin x|+|\cos x| \text { is } \frac{\pi}{2} \\ & \Rightarrow \quad I=4 \int_0^{\frac{\pi}{2}}[|\sin x|+|\cos x|] d x \\ & \text { For } x \in\left(0, \frac{\pi}{2}\right), \sin x>0 \text { and } \cos x>0 \\ & \Rightarrow \quad I=4 \int_0^{\frac{\pi}{2}}[\sin x+\cos x] d x \\ & \Rightarrow \quad[\sin x+\cos x]=1 \\ & \Rightarrow \quad I=4 \int_0^{\frac{\pi}{2}} d x=\left.4 x\right|_0 ^{\frac{\pi}{2}}=2 \pi \end{aligned} $

Given, $f(x) \cdot f(-x)=9$

$ \begin{aligned} & f(-x)= \frac{9}{f(x)} \\ & I=\int_{-23}^{23} \frac{d x}{3+f(x)} \\ & I=\int_{-23}^{23} \frac{d x}{3+f(-x)} \\ &\left\{\because \int_a^b f(x) d x=\int_a^b f(a+b-x) d x\right\} \\ & I=\int_{-23}^{23} \frac{d x}{3+\frac{9}{f(x)}} \quad \text { [from Eq. (i)] } \\ & I=\frac{1}{3} \int_{-23}^{23} \frac{f(x)}{3+f(x)} d x \end{aligned} $

Adding Eqs. (ii) and (iii), we get

$ \begin{aligned} 2 I & =\int_{-23}^{23}\left(\frac{1}{3+f(x)}+\frac{1}{3} \frac{f(x)}{3+f(x)}\right) d x \\ & =\int_{-23}^{23} \frac{3+f(x)}{3(3+f(x))} d x=\frac{1}{3} \int_{-23}^{23} d x \\ 2 I & =\frac{46}{3} \\ \Rightarrow I & =\frac{46}{6} \end{aligned} $

$ \text { } \begin{aligned} & \text { Let } I=\int_0^4 \| x-2|-x| d x \\ & I=\int_0^2|-(x-2)-x| d x+\int_2^4|x-2-x| d x \\ &=\int_0^2|2-2 x| d x+\int_2^4 2 d x \\ &=\int_0^1(2-2 x) d x+\int_1^2-(2-2 x) d x+\int_2^4 2 d x \\ &=\left[2 x-x^2\right]_0^1-\left[2 x-x^2\right]_1^2+2[x]_2^4 \\ &=[2-1-0]-[4-4-2+1]+2[4-2] \\ &=1+1+4 \\ &=6 \end{aligned} $

$\int_{-a}^a f(x) \cdot d x=\int_0^a f(x) d x+\int_0^a g(x) d x$

As we know,

$ \int_{-a}^a f(x) d x=\int_{-a}^0 f(x) d x+\int_0^a f(x) d x $

In 2nd term, on putting $x=-t$

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

Limits when $x=0, t=0$

When $x=-a, t=a$

$ \begin{aligned} \therefore \quad \int_{-a}^a f(x) d x & =\int_a^0 f(-t),(-d t)+\int_0^a f(x) d x \\ & =\int_0^a f(x) d x-\int_a^0 f(-t) d t \\ & =\int_0^a f(x)+\int_0^a f(-t) d t \\ & =\int_0^a f(x)+\int_0^a f(-x) d x \\ \therefore g(x)=f(-x) . & \end{aligned} $