Question

For $x \in\left(0, \frac{\pi}{2}\right)$, let $f_n(x)=\int n \sin 2 x\left(\sin ^{2 n-2} x-\cos ^{2 n-2} x\right) d x-\frac{1}{2^{n-1}}, n \in N$ and $f_n\left(\frac{\pi}{4}\right)=\frac{1}{2^{n-1}}$
If $m=\underset{t \rightarrow-\infty}{\text{Lim}} \left[\cot ^{-1} t\right]$, then $\int \frac{f_m(x)}{\sin ^2 x \cdot \cos ^2 x} d x$ equals
(where [] denotes greatest integer function and $c$ is constant of integration.)

• A. $\tan x-\cot x+3 x+c$
• B. $\tan x-\cot x-3 x+c$
• C. $\tan x+\cot x-3 x+c$
• D. $\tan x+\cot x+3 x+c$

Answer 1

Answer: (B)

$ f_n(x)=\int 2 n \sin x \cos x\left(\sin ^{2 n-2} x-\cos ^{2 n-2} x\right) d x-\frac{1}{2^{n-1}}$
$=\int 2 n \sin ^{2 n-1} x \cos x d x-\int 2 n \sin x \cdot \cos ^{2 n-1} x d x-\frac{1}{2^{n-1}} $
$=\sin ^{2 n} x+\cos ^{2 n} x-\frac{1}{2^{n-1}}+k$
where $k=$ constant of integration
$\Theta f_n\left(\frac{\pi}{4}\right)=\left(\frac{1}{\sqrt{2}}\right)^{2 n}+\left(\frac{1}{\sqrt{2}}\right)^{2 n}-\frac{1}{2^{n-1}}+k=\frac{1}{2^{n-1}} \Rightarrow k=\frac{1}{2^{n-1}}$ $\therefore f_n(x)=\sin ^{2 n} x+\cos ^{2 n} x$
$\Theta m =\underset{ t \rightarrow-\infty}{\text{Lim}} \left[\cot ^{-1} t \right]=[\pi]=3 $
$\therefore \text { Integration }=\int \frac{ f _3( x )}{\sin ^2 x \cos ^2 x } dx =\int \frac{\sin ^6 x +\cos ^6 x }{\sin ^2 x \cos ^2 x } dx$
$=\int \frac{\left(\sin ^2 x +\cos ^2 x \right)\left(\sin ^4 x +\cos ^4 x -\sin ^2 x \cos ^2 x \right)}{\sin ^2 x \cos ^2 x } dx$
$=\int\left(\tan ^2 x+\cot ^2 x-1\right) d x=\int\left(\sec ^2 x+\operatorname{cosec}^2 x-3\right) d x$
$=\tan x-\cot x-3 x+c (c=\text { constant of integration })$