⇒Indefinite Integration

Correct Option is (A)

Let $\mathrm{I}=\int x\sqrt{\frac{2\sin (x^2+1)-\sin 2(x^2+1)}{2\sin (x^2+1)+\sin 2(x^2+1)}}dx$

$=\int x\sqrt{\frac{2\sin (x^2+1)-2\sin (x^2+1)\cos (x^2+1)}{2\sin x(x^2+1)+2\sin (x^2+1)\cos (x^2+1)}}\mathrm{d}x$

$=\int x\sqrt{\frac{1-\cos (x^2+1)}{1+\cos (x^2+1)}}\mathrm{d}x$

$=\int x\sqrt{\frac{2{\sin }^2\left(\frac{x^2+1}{2}\right)}{2{\cos }^2\left(\frac{x^2+1}{2}\right)}}dx$

$=\int x\tan \left(\frac{x^2+1}{2}\right)dx$

Let $\left(\frac{x^2+1}{2}\right)=\mathrm{t}\Rightarrow x\mathrm{d}x=\mathrm{dt}$

$\begin{array}{rl}\therefore \mathrm{I}& =\int \tan \mathrm{t}d\mathrm{t}\\ & =\log (\sec \mathrm{t})+\mathrm{c}\\ & =\log (\sec \left(\frac{x^2+1}{2}\right))+\mathrm{c}\end{array}$

Correct Option is (D)

$\begin{array}{rl}& I=\int \frac{\cos 8x+1}{\cot 2x-\tan 2x}dx\\ & =\int \frac{2{\cos }^2\left(\frac{8x}{2}\right)}{\frac{\cos 2x}{\sin 2x}-\frac{\sin 2x}{\cos 2x}}dx\\ & \dots [\because 1+\cos \theta =2{\cos }^2\frac{\theta }{2}]\end{array}$

$\begin{array}{rl}& =\int \frac{2{\cos }^2(4x)\times \sin 2x\times \cos 2x}{{\cos }^{2}2x-{\sin }^{2}2x}\mathrm{d}x\\ & =\int \frac{{\cos }^2(4x)\sin (4x)}{\cos (4x)}dx\\ & \dots \left[\begin{array}{c}\because \sin 2\theta =2\sin \theta \cos \theta \text{ and }\\ \cos 2\theta ={\cos }^2\theta -{\sin }^2\theta \end{array}\right]\end{array}$

$\begin{array}{rl}& =\frac{1}{2}\int 2\sin (4x)\cos (4x)dx\\ & =\frac{1}{2}\int \sin 8xdx\\ & =\frac{-\cos 8x}{2\times 8}+c\\ & =\frac{-\cos 8x}{16}+c\end{array}$

Comparing with '$A\cos 8x+c$', we get $A=\frac{-1}{16}$

Correct Option is (B)

Let

$\begin{array}{rl}\mathrm{I}& =\int (1-\cos x)\cdot {\mathrm{cosec}}^2\mathrm{d}x\\ & =\int \frac{2{\sin }^2\frac{x}{2}}{[\sin x{]}^2}\mathrm{d}x\dots [\because 1-\cos \theta =2{\sin }^2\frac{\theta }{2}]\\ & =\int \frac{2{\sin }^2\frac{x}{2}}{{[2\sin \frac{x}{2}\cos \frac{x}{2}]}^2}\mathrm{d}x\dots [\because \sin \theta =2\sin \frac{\theta }{2}\cos \frac{\theta }{2}]\\ & =\frac{1}{2}\int {\sec }^2\frac{x}{2}\mathrm{d}x\\ & =\tan \frac{x}{2}+\mathrm{c}\end{array}$

Correct Option is (C)

$\begin{array}{rl}& \text{ Let }\mathrm{I}=\int \frac{(x^2-1)\mathrm{d}x}{x^3\sqrt{2x^4-2x^2+1}}\\ & =\int \frac{(x^2-1)\mathrm{d}x}{x^3\cdot x^2\sqrt{(2-\frac{2}{x^2}+\frac{1}{x^4})}}\\ & =\int \frac{\left(\frac{x^2-1}{x^5}\right)\mathrm{d}x}{\sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}}\\ & \text{ Put }2-\frac{2}{x^2}+\frac{1}{x^4}=\mathrm{t}\\ & \Rightarrow (\frac{4}{x^3}-\frac{4}{x^5})\mathrm{d}x=\mathrm{dt}\\ & \Rightarrow \frac{x^2-1}{x^5}\mathrm{d}x=\frac{\mathrm{dt}}{4}\\ & \therefore I=\int \frac{\frac{dt}{4}}{\sqrt{t}}\\ & =\frac{1}{4}\int {\mathrm{t}}^{\frac{-1}{2}}\mathrm{dt}\\ & =\frac{1}{4}\cdot \frac{t^{\frac{1}{2}}}{\frac{1}{2}}+c\\ \therefore & \mathrm{I}=\frac{1}{2}\sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}+\mathrm{c}\end{array}$

Correct Option is (B)

Let $I=\int \frac{dx}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}$

Put $x^{\frac{1}{6}}=t\Rightarrow x=t^6\Rightarrow dx=6t^{5}dt$. Also $x^{\frac{1}{2}}=t^3$ and $x^{\frac{1}{3}}=t^2$

$\begin{array}{rl}\therefore I& =6\int \frac{t^{5}dt}{t^2(1+t)}\\ & =6\int \frac{t^3}{(1+t)}dt=6\int \frac{(t^3+1)-1}{(1+t)}dt=6\int \frac{(t+1)(t^2-t+1)}{(1+t)}dt-6\int \frac{dt}{1+t}\\ & =6\int (t^2-t+1)dt-6\log |1+t|+c=\frac{6t^3}{3}-\frac{6t^2}{2}+6t-6[\log |1+t|]+c\\ & =2\sqrt{x}-3\sqrt[3]{x}+6\sqrt[6]{x}-6\log |1+\sqrt[6]{x}|+c\end{array}$