Question

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

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

Answer 1

Answer: (B)

$f_n(x)=\int 2n\sin x\cos x\left({\sin }^{2n-2}x-{\cos }^{2n-2}x\right)dx-\frac{1}{2^{n-1}}$
$=\int 2n{\sin }^{2n-1}x\cos xdx-\int 2n\sin x\cdot {\cos }^{2n-1}xdx-\frac{1}{2^{n-1}}$
$={\sin }^{2n}x+{\cos }^{2n}x-\frac{1}{2^{n-1}}+k$
where $k=$ constant of integration
$\Theta f_n\left(\frac{\pi }{4}\right)={\left(\frac{1}{\sqrt{2}}\right)}^{2n}+{\left(\frac{1}{\sqrt{2}}\right)}^{2n}-\frac{1}{2^{n-1}}+k=\frac{1}{2^{n-1}}\Rightarrow k=\frac{1}{2^{n-1}}$ $\therefore f_n(x)={\sin }^{2n}x+{\cos }^{2n}x$
$\Theta m=\underset{t\to -\infty }{\mathrm{Lim}}\left[{\cot }^{-1}t\right]=[\pi ]=3$
$\therefore \text{ Integration }=\int \frac{f_3(x)}{{\sin }^{2}x{\cos }^{2}x}dx=\int \frac{{\sin }^{6}x+{\cos }^{6}x}{{\sin }^{2}x{\cos }^{2}x}dx$
$=\int \frac{\left({\sin }^{2}x+{\cos }^{2}x\right)\left({\sin }^{4}x+{\cos }^{4}x-{\sin }^{2}x{\cos }^{2}x\right)}{{\sin }^{2}x{\cos }^{2}x}dx$
$=\int \left({\tan }^{2}x+{\cot }^{2}x-1\right)dx=\int \left({\sec }^{2}x+{\mathrm{cosec}}^{2}x-3\right)dx$
$=\tan x-\cot x-3x+c(c=\text{ constant of integration })$