Question

$\begin{array}{cccc}& \text{List I}& & \text{List II}\\ \text{(A)}& \text{If}\int \left(\frac{x^2+{\cos }^{2}x}{1+x^2}\right){\text{cosec}}^{2}xdx=m{\cot }^{-1}x& & \\ & +\frac{n\text{cosec}x}{\sec x},\text{then}& \text{(P)}& m=1\\ & & & \\ \text{(B)}& \text{If}\int \sqrt{x+\sqrt{x^2+2}}dx=\frac{m}{3}(x+\sqrt{x^2+2}{)}^{3/2}& \text{(Q)}& n=-1\\ & -\frac{n}{\sqrt{x+\sqrt{x^2+2}}},\text{then}& & \\ \text{(C)}& \text{If}\int \frac{x{\tan }^{-1}x}{(1+x^2{)}^{3/2}}dx=\frac{mx}{\sqrt{1+x^2}}& \text{(R)}& n=2\\ & +\frac{n{\tan }^{-1}x}{\sqrt{1+x^2}},\text{then}& & \\ \text{(D)}& \text{If}\int \frac{\sin 2x}{{\sin }^{4}x+{\cos }^{4}x}dx=n{\cot }^{-1}\left({\tan }^{2}x\right)& & \\ & +m{\sin }^{2}x,\text{then}& \text{(S)}& m=-1\\ & & \text{(T)}& m=0\\ & & \text{(U)}& n=1\end{array}$

Which of the following is the only INCORRECT combination?

• A. $\left(A\right)\to \left(P\right),\left(Q\right)$
• B. $\left(B\right)\to \left(P\right),\left(R\right)$
• C. $\left(C\right)\to \left(S\right),\left(R\right)$
• D. $\left(D\right)\to \left(T\right),\left(Q\right)$

Answer 1

The correct option is C $\left(C\right)\to \left(S\right),\left(R\right)$

$\left(A\right)$
$I=\int \left(\frac{x^2+{\cos }^{2}x}{1+x^2}\right){\text{cosec}}^{2}xdx$
$=\int \left(\frac{x^2+1-{\sin }^{2}x}{1+x^2}\right){\text{cosec}}^{2}xdx$
$=\int ({\text{cosec}}^{2}x-\frac{1}{1+x^2})dx$
$=-\cot x+{\cot }^{-1}x+k$
$=-\frac{\text{cosec}x}{\sec x}+{\cot }^{-1}x+k$
$\therefore m=1,n=-1$
$\left(A\right)\to \left(P\right),\left(Q\right)$

$\left(B\right)$
$I=\int \sqrt{x+\sqrt{x^2+2}}dx$
Put $\sqrt{x^2+2}+x=t$
$\Rightarrow \sqrt{x^2+2}-x=\frac{2}{t}$
$\therefore 2x=t-\frac{2}{t}\Rightarrow 2dx=(1+\frac{2}{t^2})dt$
$I=\frac{1}{2}\int (\sqrt{t}+\frac{2}{t^{3/2}})dt$
$\Rightarrow I=\frac{1}{2}\frac{t^{3/2}}{3/2}+\frac{1}{-\frac{1}{2}\sqrt{t}}+k$
$=\frac{1}{3}(x+\sqrt{x^2+2}{)}^{3/2}-\frac{2}{\sqrt{x+\sqrt{x^2+2}}}+k$
$\therefore m=1,n=2$
$\left(B\right)\to \left(P\right),\left(R\right)$

$\left(C\right)$
$I=\int \frac{x{\tan }^{-1}x}{(1+x^2{)}^{3/2}}dx=\int \frac{x{\tan }^{-1}x}{(1+x^2)\sqrt{1+x^2}}dx$
Put $x=\tan \theta$, where $\theta \in (\frac{-\pi }{2},\frac{\pi }{2})$
$\Rightarrow dx={\sec }^2\theta d\theta$
$\Rightarrow \frac{dx}{1+x^2}=d\theta$
So, $I=\int \frac{\theta \tan \theta }{\sqrt{1+{\tan }^2\theta }}d\theta$
$=\int \frac{\theta \tan \theta }{\sec \theta }d\theta$
$=\int \theta \sin \theta d\theta$
$=-\theta \cos \theta +\int \cos \theta =-\theta \cos \theta +\sin \theta +k$
Now, $x=\tan \theta$
$\Rightarrow \cos \theta =\frac{1}{\sqrt{1+x^2}}$ and $\sin \theta =\frac{x}{\sqrt{1+x^2}}$
$\therefore I=\frac{x-{\tan }^{-1}x}{\sqrt{1+x^2}}$
Hence, $m=1,n=-1$
$\left(C\right)\to \left(P\right),\left(Q\right)$

$\left(D\right)$
$\int \frac{\sin 2x}{{\sin }^{4}x+{\cos }^{4}x}dx$
Dividing numerator and denominator by ${\cos }^{4}x$,
$I=\int \frac{2\tan x{\sec }^{2}x}{{\tan }^{4}x+1}dx$
Put ${\tan }^{2}x=t\Rightarrow 2\tan x{\sec }^{2}xdx=dt$
$\therefore I=\int \frac{dt}{t^2+1}$
$={\tan }^{-1}\left(t\right)+k_1$
$=-{\cot }^{-1}\left(t\right)+k$, where $k=\frac{\pi }{2}+k_1$
$=-{\cot }^{-1}\left({\tan }^{2}x\right)+k$
$\therefore m=0,n=-1$
$\left(D\right)\to \left(T\right),\left(Q\right)$