Definite Integration

$ \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} $

$\int_0^1 {\left[ {2x - |3{x^2} - 5x + 2| + 1} \right]dx} $

$3{x^2} - 5x + 2 = 0$

$ \Rightarrow 3{x^2} - 3x - 2x + 2 = 0$

$ \Rightarrow 3x(x - 1) - 2(x - 1) = 0$

$ \Rightarrow (x - 1)(3x - 2) = 0$

$\therefore$ In 0 to ${2 \over 3},\,|3{x^2} - 5x + 2| = 3{x^2} - 5x + 2$

And in ${2 \over 3}$ to 1, $|3{x^2} - 5x + 2| = - (3{x^2} - 5x + 2)$

$\therefore$ $\int_0^{{2 \over 3}} {\left[ {2x - 3{x^2} + 5x - 2 + 1} \right]dx + \int_{{2 \over 3}}^1 {\left[ {2x + 3{x^2} - 5x + 2 + 1} \right]dx} } $

$ = \int_0^{{2 \over 3}} {\left[ { - 3{x^2} + 7x - 1} \right]dx + \int_{{2 \over 3}}^1 {\left[ {3{x^2} - 3x + 3} \right]dx} } $

Now, let $f(x) = - 3{x^2} + 7x - 1$

$ \Rightarrow f'(x) = - 6x + 7$

$ = - 6\left( {x - {7 \over 6}} \right)$

$f(0) = - 1$

$f\left( {{2 \over 3}} \right) = - 3 \times {4 \over 9} + {{14} \over 3} - 1$

$ = {7 \over 3} = 2.33$

$\therefore$ Range of $f(x) = \left[ { - 1,{7 \over 3}} \right]$

Integer value between $ - 1$ to ${7 \over 3} = - 1,0,1,2,{7 \over 3}$

When $f(x) = - 1$

$ \Rightarrow - 3{x^2} + 7x - 1 = - 1$

$ \Rightarrow 3{x^2} - 7x = 0$

$ \Rightarrow x(3x - 7) = 0$

$ \Rightarrow x = 0,{7 \over 3}$ (not possible as ${7 \over 3} \notin (0,2)$)

When $f(x) = 0$

$ \Rightarrow - 3{x^2} + 7x - 1 = 0$

$ \Rightarrow 3{x^2} - 7x + 1 = 0$

$ \Rightarrow x = {{7\, \pm \,\sqrt {49 - 12} } \over 6}$

$ = {{7\, \pm \sqrt {37} } \over 6}$

$\therefore$ $x = {{7 - \sqrt {37} } \over 6}$

When $f(x) = 1$

$ \Rightarrow - 3{x^2} + 7x - 1 = 1$

$ \Rightarrow 3{x^2} - 7x + 2 = 0$

$ \Rightarrow x = {{7\, \pm \,\sqrt {49 - 24} } \over 6}$

$ = {{7\, \pm \,\sqrt {25} } \over 6}$

$ = {{7\, \pm \,5} \over 6}$

$ \Rightarrow x = 2,\,{1 \over 3}$

When $f(x) = 2$

$ \Rightarrow - 3{x^2} + 7x - 1 = 2$

$ \Rightarrow 3{x^2} - 7x + 3 = 0$

$ \Rightarrow x = {{7\, \pm \,\sqrt {49 - 36} } \over 6}$

$ = {{7\, \pm \,\sqrt {13} } \over 6}$

$\therefore$ $x = {{7\, - \sqrt {13} } \over 6}$

$\therefore$ Possible values of $x = 0,\,{{7\, - \,\sqrt {37} } \over 6},{1 \over 3},{{7\, - \,\sqrt {13} } \over 6},{2 \over 3}$

$\therefore$ $\int_0^{{2 \over 3}} {\left[ { - 3{x^2} + 7x - 1} \right]dx} $

$ = \int_0^{{{7\, - \,\sqrt {37} } \over 6}} {( - 1)dx + \int_{{{7\, - \,\sqrt {37} } \over 6}}^{{1 \over 3}} {0\,dx + \int_{{1 \over 3}}^{{{7\, - \,\sqrt {13} } \over 6}} {(1)\,dx + \int_{{{7\, - \,\sqrt {13} } \over 6}}^{{2 \over 3}} {2\,dx} } } } $

$ = {{ - 7\, + \,\sqrt {37} } \over 6} + 0 + {{7\, - \,\sqrt {13} } \over 6} - {1 \over 3} + 2\left[ {{2 \over 3} - {{7\, - \,\sqrt {13} } \over 6}} \right]$

$ = {{ - 7 + \sqrt {37} + 7 - \sqrt {13} - 2 + 8 - 14 + 2\sqrt {13} } \over 6}$

$ = {{\sqrt {37} + \sqrt {13} - 8} \over 6}$

Now, $\int_{{2 \over 3}}^1 {\left[ {3{x^2} - 3x + 3} \right]dx} $

Let $f(x) = 3{x^2} - 3x + 3$

$f'(x) = 6x - 3 = 3(2x - 1)$

$f'(x) = 0 \Rightarrow 2x - 1 = 0 \Rightarrow x = {1 \over 2}$

$f\left( {{2 \over 3}} \right) = 3\left( {{4 \over 9} - {2 \over 3} + 1} \right)$

$ = {7 \over 3} = 2.3$

$f(1) = 3(1 - 1 + 1)$

$ = 3$

No integer values present in between ${7 \over 3}$ and 3.

$\therefore$ Possible value of $x = {2 \over 3},\,1$

So, integration don't break anywhere.

$\therefore$ $\int_{{2 \over 3}}^1 {\left[ {3{x^2} - 3x + 3} \right]dx} $

$ = \int_{{2 \over 3}}^1 {\left[ {2.33} \right]\,dx} $

$ = \int_{{2 \over 3}}^1 {2\,dx = 2\left( {1 - {2 \over 3}} \right) = {2 \over 3}} $

$\therefore$ Value of Integration

$ = {{\sqrt {37} + \sqrt {13} - 1} \over 6} + {2 \over 3}$

$ = {{\sqrt {37} + \sqrt {13} - 8 + 4} \over 6}$

$ = {{\sqrt {37} + \sqrt {13} - 4} \over 6}$

$I = \int\limits_0^{\pi /2} {{1 \over {3 + 2\sin x + \cos x}}dx} $

$ = \int\limits_0^{\pi /2} {{{(1 + {{\tan }^2}x/2)dx} \over {3(1 + {{\tan }^2}x/2) + 2(2\tan x/2) + (1 - {{\tan }^2}x/2)}}} $

Let $\tan x/2 = t \Rightarrow {\sec ^2}x/2dx = 2dt$

$I = \int\limits_0^1 {{{2dt} \over {4 + 2{t^2} + 4t}}} $

$ = \int\limits_0^1 {{{dt} \over {{t^2} + 2t + 2}} = \int\limits_0^1 {{{dt} \over {{{(t + 1)}^2} + 1}}} } $

$ = \left. {{{\tan }^{ - 1}}(t + 1)} \right|_0^1 = {\tan ^{ - 1}}2 - {\pi \over 4}$

$f(\alpha ) = \int_1^\alpha {{{{{\log }_{10}}t} \over {1 + t}}dt} $ ...... (i)

$f\left( {{1 \over \alpha }} \right) = \int_1^{{1 \over \alpha }} {{{{{\log }_{10}}t} \over {1 + t}}dt} $

Substituting $t \to {1 \over p}$

$f\left( {{1 \over \alpha }} \right) = \int_1^\alpha {{{{{\log }_{10}}\left( {{1 \over p}} \right)} \over {1 + {1 \over p}}}\left( {{{ - 1} \over {{p^2}}}} \right)dp} $

$ = \int_1^\alpha {{{{{\log }_{10}}p} \over {p(p + 1)}}dp = \int_1^\alpha {\left( {{{{{\log }_{10}}t} \over t} - {{{{\log }_{10}}t} \over {t + 1}}} \right)dt} } $ ....... (ii)

By (i) + (ii)

$f(\alpha ) + f\left( {{1 \over \alpha }} \right) = \int_1^\alpha {{{{{\log }_{10}}t} \over t}dt = \int_1^\alpha {{{\ln t} \over t}\,.\,{{\log }_{10}}e\,dt} } $

$ = {{{{(\ln \alpha )}^2}} \over {2{{\log }_e}10}}$

$\alpha = {e^3} \Rightarrow f({e^3}) + f({e^{ - 3}}) = {9 \over {2{{\log }_e}10}}$

${I_n}(x) = \int\limits_0^x {{1 \over {{{({t^2} + 5)}^n}}}dt} $

$ = \int\limits_0^x {{1 \over {\underbrace {{{({t^2} + 5)}^n}}_I}} \times \mathop I\limits_{II} \,dt} $

$ = \left. {{t \over {{{({t^2} + 5)}^n}}}} \right|_0^x - \int\limits_0^x {{{ - 2nt} \over {{{({t^2} + 5)}^{n + 1}}}} \times t\,dt} $

$ = {x \over {{{({x^2} + 5)}^n}}} + \int\limits_0^x {2n\left( {{{{t^2} + 5 - 5} \over {{{({t^2} + 5)}^{n + 1}}}}} \right)dt} $

${I_n}(x) = {x \over {{{({x^2} + 5)}^n}}} + 2n\,{I_n}(x) - 10n\,{I_{n + 1}}(x)$

$10n\,{I_{n + 1}}(x) - (2n - 1)\,{I_n}(x) = xI{'_n}(x)$

For $n = 5$

$50{I_6}(x) - 9{I_5}(x) = xI{'_5}(x)$

$f(x) = \int\limits_0^x {{e^{x - t}}f'(t)dt - ({x^2} - x + 1){e^x}} $

$f(x) = {e^x}\int\limits_0^x {{e^{ - t}}f'(t)dt - ({x^2} - x + 1){e^x}} $

${e^{ - x}}f(x) = \int\limits_0^x {{e^{ - t}}f'(t)dt - ({x^2} - x + 1)} $

Differentiate on both side

${e^{ - x}}f'(x) + ( - f(x){e^{ - x}}) = {e^{ - x}}f'(x) - 2x + 1$

$f(x) = {e^x}(2x - 1)$

$f'(x) = {e^x}(2) + {e^x}(2x - 1)$

$ = {e^x}(2x + 1)$

$x = - {1 \over 2}$

$f''(x) = {e^x}(2) + (2x + 1){e^x}$

$ = {e^x}(2x + 3)$

For $x = - {1 \over 2}\,\,f''(x) > 0$

$\Rightarrow$ Maxima

$\therefore$ Max. $ = {e^{ - {1 \over 2}}}( - 1 - 1)$

$\therefore$ $ - {2 \over {\sqrt e }}$

$f(x) = 2 + |x| - |x - 1| + |x + 1|,\,x \in R$

$\therefore$ $f(x) = \left\{ {\matrix{ { - x} & , & {x < - 1} \cr {x + 2} & , & { - 1 \le x < 0} \cr {3x + 2} & , & {0 \le x < 1} \cr {x + 4} & , & {x \ge 1} \cr } } \right.$

$\therefore$ $f'\left( { - {3 \over 2}} \right) + f'\left( { - {1 \over 2}} \right) + f'\left( {{1 \over 2}} \right) + f'\left( {{3 \over 2}} \right) = - 1 + 1 + 3 + 1 = 4$

and $\int\limits_{ - 2}^2 {f(x)dx = \int\limits_{ - 2}^{ - 1} {f(x)dx + \int\limits_{ - 1}^0 {f(x)dx + \int\limits_0^1 {f(x)dx + \int\limits_1^2 {f(x)dx} } } } } $

$ = \left[ { - {{{x^2}} \over 2}} \right]_2^{ - 1} + \left[ {{{{{(x + 2)}^2}} \over 2}} \right]_{ - 1}^0 + \left[ {{{{{(3x + 2)}^2}} \over 6}} \right]_0^1 + \left[ {{{{{(x + 4)}^2}} \over 2}} \right]_1^2$

$ = {3 \over 2} + {3 \over 2} + {7 \over 2} + {{11} \over 2} = {{24} \over 2} = 12$

$\therefore$ Only (S2) is correct

$\int\limits_0^2 {|2{x^2} - 3x|dx + \int\limits_0^2 {\left[ {x - {1 \over 2}} \right]dx} } $

$ = \int\limits_0^{3/2} {(3x - 2{x^2})dx + \int\limits_{3/2}^2 {(2{x^2} - 3x)dx + \int\limits_0^{1/2} { - 1dx + \int\limits_{1/2}^{3/2} {0\,dx + \int\limits_{3/2}^2 {1dx} } } } } $

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

$ = \left( {{{27} \over 8} - {{27} \over {12}}} \right) + \left( {{{16} \over 3} - 6 - {{27} \over {12}} + {{27} \over 8}} \right)$

$ = {{19} \over {12}}$

$f(x) = a\sin \left( {{{\pi [x]} \over 2}} \right) + [2 - x]\,a \in R$

Now,

$\because$ $\mathop {\lim }\limits_{x \to - 1} f(x)$ exist

$\therefore$ $\mathop {\lim }\limits_{x \to - {1^ - }} f(x) = \mathop {\lim }\limits_{x \to - {1^ + }} f(x)$

$ \Rightarrow a\sin \left( {{{ - 2\pi } \over 2}} \right) + 3 = a\sin \left( {{{ - \pi } \over 2}} \right) + 2$

$ \Rightarrow - a = 1 \Rightarrow a = - 1$

Now, $\int_0^4 {f(x)dx = \int_0^4 {\left( { - \sin \left( {{{\pi [x]} \over 2}} \right) + [2 - x]} \right)dx} } $

$ = \int_0^1 {1dx + \int_1^2 { - 1dx + \int_2^3 { - 1dx + \int_3^4 {(1 - 2)dx} } } } $

$ = 1 - 1 - 1 - 1 = - 2$

I comes out around 1.536 which is not satisfied by any given options.

$\int\limits_{\pi /4}^{\pi /3} {{{8x - 2x} \over x}dx > I > \int\limits_{\pi /4}^{\pi /3} {{{8\sin x - 2x} \over x}dx} } $

${\pi \over 2} > I > \int\limits_{\pi /4}^{\pi /3} {\left( {{{8\sin x} \over x} - 2} \right)dx} $

${{\sin x} \over x}$ is decreasing in $\left( {{\pi \over 4},{\pi \over 3}} \right)$ so it attains maximum at $x = {\pi \over 4}$

$I > \int\limits_{\pi /4}^{\pi /3} {\left( {{{8\sin x/3} \over {x/3}} - 2} \right)dx} $

$I > \sqrt 3 - {\pi \over 6}$

$\because$ f(x) is continuous at x = 4

$ \Rightarrow f({4^ - }) = f({4^ + })$

$ \Rightarrow 16 + 4b = \int\limits_0^4 {(5 - |t - 3|)dt} $

$ = \int\limits_0^3 {(2 + t)dt + \int\limits_3^4 {(8 - t)dt} } $

$ = \left. {2t + {{{t^2}} \over 2}} \right)_0^3 + \left. {8t - {{{t^2}} \over 3}} \right]_3^4$

$ = 6 + {9 \over 2} - 0 + (32 - 8) - \left( {24 - {9 \over 2}} \right)$

$16 + 4b = 15$

$ \Rightarrow b = {{ - 1} \over 4}$

$ \Rightarrow f(x) = \left\{ {\matrix{ {\int\limits_0^x {5 - |t - 3|\,dt} } & {x > 4} \cr {{x^2} - {x \over 4}} & {x \le 4} \cr } } \right.$

$ \Rightarrow f'(x) = \left\{ {\matrix{ {5 - |x - 3|} & {x > 4} \cr {2x - {1 \over 4}} & {x \le 4} \cr } } \right.$

$ \Rightarrow f'(x) = \left\{ {\matrix{ {8 - x} & {x > 4} \cr {2x - {1 \over 4}} & {x \le 4} \cr } } \right.$

$f'(x) < 0 = x \in \left( { - \infty ,{1 \over 8}} \right) \cup (8,\infty )$

$f'(3) + f'(5) = 6 - {1 \over 4} = {{35} \over 4}$

$f'(x) = 0 \Rightarrow x = {1 \over 8}$ have local minima

$\therefore$ (C) is only incorrect option.

$I = \int\limits_0^{20\pi } {{{\left( {|\sin x| + |\cos x|} \right)}^2}\,dx} $

$ = 20\int\limits_0^\pi {\left( {1 + |\sin 2x|} \right)\,dx} $

$ = 40\int\limits_0^{{\pi \over 2}} {(1 + \sin 2x)\,dx} $

$ = \left. {40\left( {x - {{\cos 2x} \over 2}} \right)} \right|_0^{{\pi \over 2}}$

$ = 40\left( {{\pi \over 2} + {1 \over 2} + {1 \over 2}} \right) = 20(\pi + 2)$

$a = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {{{2n} \over {{n^2} + {k^2}}}} $

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

$a = \int\limits_0^1 {{2 \over {1 + {x^2}}}dx = 2{{\tan }^{ - 1}}x\int_0^1 { = {\pi \over 2}} } $

$f(x) = \sqrt {{{1 - \cos x} \over {1 + \cos x}}} ,\,x \in (0,1)$

$f(x) = {{1 - \cos x} \over {\sin x}} = \cos ec\,x - \cot x$

$f'(x) = \cos e{c^2}x - \cos ec\,x\cot x$

$\left. {\matrix{ {f\left( {{a \over 2}} \right) = f\left( {{\pi \over 4}} \right) = \sqrt 2 - 1} \cr {f'\left( {{a \over 2}} \right) = f'\left( {{\pi \over 4}} \right) = 2 - \sqrt 2 } \cr } } \right\}f'\left( {{a \over 2}} \right) = \sqrt 2 \,.\,f\left( {{a \over 2}} \right)$