⇒Indefinite Integration

Correct Option is (A)

$\begin{array}{rl}\text{Let}I& =\int \frac{1}{{\cos }^{3}x\sqrt{\sin 2x}}dx\\ & =\frac{1}{\sqrt{2}}\int \frac{{\sec }^{3}x}{\sqrt{\sin x\cos x}}d\\ & =\frac{1}{\sqrt{2}}\int \frac{{\sec }^{4}x}{\sqrt{\tan x}}dx\end{array}$

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

$\begin{array}{rl}\therefore I& =\frac{1}{\sqrt{2}}\int \frac{1+t^2}{\sqrt{t}}dt\\ & =\frac{1}{\sqrt{2}}\int t^{\frac{1}{2}}dt+\frac{1}{\sqrt{2}}\int t^{\frac{3}{2}}dt\\ & =\frac{1}{\sqrt{2}}\left(\frac{t^{\frac{1}{2}}}{\frac{1}{2}}\right)+\frac{1}{\sqrt{2}}\left(\frac{t^{\frac{5}{2}}}{\frac{5}{2}}\right)+c\\ & =\sqrt{2}(\sqrt{\tan x}+\frac{1}{5}(\tan x{)}^{\frac{5}{2}})+c\end{array}$

Correct Option is (C)

$\begin{array}{rl}\text{ Let }\mathrm{I}& =\int \frac{\sqrt{1-x^2}}{x^4}dx\\ & =\int \frac{x\sqrt{\frac{1}{x^2}-1}}{x^4}dx\\ & =\int \frac{\sqrt{\frac{1}{x^2}-1}}{x^3}dx\end{array}$

Let $\frac{1}{x^2}-1=\mathrm{t}$

$\begin{array}{rl}& \therefore \frac{-2}{x^3}\mathrm{d}x=\mathrm{dt}\Rightarrow \frac{1}{x^3}\mathrm{d}x=\frac{-\mathrm{dt}}{2}\\ & \therefore \mathrm{I}=-\frac{1}{2}\int \sqrt{\mathrm{t}}\mathrm{dt}\\ & =\frac{-1}{2}\times \frac{(\mathrm{t}{)}^{\frac{3}{2}}}{\frac{3}{2}}+\mathrm{c}\\ & =\frac{-1}{3}\times {(\frac{1}{x^2}-1)}^{\frac{3}{2}}+\mathrm{c}\end{array}$

$\begin{array}{rl}& =\frac{-1}{3}\times \frac{{(1-x^2)}^{\frac{3}{2}}}{{\left(x^2\right)}^{\frac{3}{2}}}+c\\ & =\frac{-1}{3}\times \frac{{\left(\sqrt{1-x^2}\right)}^3}{x^3}\end{array}$

Comparing with $\mathrm{A}(x){\left(\sqrt{1-x^2}\right)}^{\mathrm{m}}+\mathrm{c}$, we get $A(x)=\frac{-1}{3x^3}$ and $\mathrm{m}=3$

$\therefore (\mathrm{A}(x){)}^{\mathrm{m}}={\left(\frac{-1}{3x^3}\right)}^3=\frac{-1}{27x^9}$

Correct Option is (B)

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

Correct Option is (A)

Let $I=\int \frac{\mathrm{cosec}xdx}{{\cos }^2(1+\log \tan \frac{x}{2})}dx$

Let $1+\log (\tan \frac{x}{2})=\mathrm{t}$

Differentiating both sides w.r.t. $\mathrm{t}$, we get

$\frac{1}{\tan \frac{x}{2}}{\sec }^2\frac{x}{2}\times \frac{1}{2}\mathrm{d}x=\mathrm{dt}$

$\therefore \frac{1}{2\sin \frac{x}{2}\cos \frac{x}{2}}\mathrm{d}x=\mathrm{dt}$

$\therefore \mathrm{cosec}x\mathrm{d}x=\mathrm{dt}$

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

Correct Option is (B)

Let

$\begin{array}{rl}& I=\int \frac{{\sin }^{2}x{\cos }^{2}x}{{({\sin }^{5}x+{\cos }^{3}x{\sin }^{2}x+{\sin }^{3}x{\cos }^{2}x+{\cos }^{5}x)}^2}\mathrm{d}x\\ & =\int \frac{{\sin }^{2}x{\cos }^{2}x}{{({\sin }^{5}x+{\sin }^{3}x{\cos }^{2}x+{\cos }^{3}x{\sin }^{2}x+{\cos }^{5}x)}^2}\mathrm{d}x\\ & =\int \frac{{\sin }^{2}x{\cos }^{2}x}{{[{\sin }^{3}x({\sin }^{2}x+{\cos }^{2}x)+{\cos }^{3}x({\sin }^{2}x+{\cos }^{2}x)]}^2}\mathrm{d}x\\ & =\int \frac{{\sin }^{2}x{\cos }^{2}x}{{({\sin }^{3}x+{\cos }^{3}x)}^2}\mathrm{d}x\\ & =\int \frac{{\sec }^{2}x{\tan }^{2}x}{{(1+{\tan }^{3}x)}^2}\mathrm{d}x\end{array}$

....[Dividing numerator and denominator by ${\cos }^{6}x$ ]

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

Differentiating w.r.t. $x$, we get

$\begin{array}{rl}& 3{\tan }^{2}x{\sec }^{2}x\mathrm{d}x=\mathrm{dt}\\ \therefore & {\tan }^{2}x{\sec }^{2}x\mathrm{d}x=\frac{1}{3}\mathrm{dt}\end{array}$

$\begin{array}{rl}\therefore \mathrm{I}& =\frac{1}{3}\int \frac{1}{{\mathrm{t}}^2}\mathrm{dt}\\ & =\frac{-1}{3\mathrm{t}}+\mathrm{c}\\ & =\frac{-1}{3(1+{\tan }^{3}x)}+\mathrm{c}\end{array}$