Question

Let $f$ be a function defined as $f\left(x\right)=\frac{1}{1+(\tan x{)}^m},$ where $m$ is a constant. Then

• A. $\underset{0}{\overset{\pi /4}{\int }}\frac{xf\left(x\right)}{{\cos }^{2}x}dx=\frac{\pi }{4}\ln 2,$ when $m=1$
• B. $\underset{0}{\overset{\pi /4}{\int }}\frac{xf\left(x\right)}{{\cos }^{2}x}dx=\frac{\pi }{8}\ln 2,$ when $m=1$
• C. $\underset{0}{\overset{\pi /2}{\int }}f\left(x\right)dx=\frac{\pi }{4},$ when $m=\sqrt{2}$
• D. $\underset{0}{\overset{\pi /2}{\int }}f\left(x\right)dx=\frac{\pi }{2},$ when $m=\sqrt{2}$

Answer 1

Answer: (B), (C)

The correct options are
B $\underset{0}{\overset{\pi /4}{\int }}\frac{xf\left(x\right)}{{\cos }^{2}x}dx=\frac{\pi }{8}\ln 2,$ when $m=1$
C $\underset{0}{\overset{\pi /2}{\int }}f\left(x\right)dx=\frac{\pi }{4},$ when $m=\sqrt{2}$

For $m=1$
$I=\underset{0}{\overset{\pi /4}{\int }}\frac{xf\left(x\right)}{{\cos }^{2}x}dx$
$=\underset{0}{\overset{\pi /4}{\int }}\frac{x}{(\sin x+\cos x)\cos x}dx$

$\sin x+\cos x=\sqrt{2}\cos (\frac{\pi }{4}-x)$

$I=\underset{0}{\overset{\pi /4}{\int }}\frac{x}{\sqrt{2}\cos (\frac{\pi }{4}-x)\cos x}dx...\left(1\right)$

Using $\underset{0}{\overset{a}{\int }}f\left(x\right)dx=\underset{0}{\overset{a}{\int }}f(a-x)dx$

$I=\underset{0}{\overset{\pi /4}{\int }}\frac{\frac{\pi }{4}-x}{\sqrt{2}\cos (\frac{\pi }{4}-x)\cos x}dx...\left(2\right)$

Adding eqn $\left(1\right)$ and $\left(2\right)$, we get

$2I=\underset{0}{\overset{\pi /4}{\int }}\frac{\frac{\pi }{4}}{\sqrt{2}\cos (\frac{\pi }{4}-x)\cos x}dx$

$I=\frac{\pi }{8}\underset{0}{\overset{\pi /4}{\int }}\frac{\frac{1}{\sqrt{2}}}{\cos (\frac{\pi }{4}-x)\cos x}dx$

$I=\frac{\pi }{8}\underset{0}{\overset{\pi /4}{\int }}\frac{\sin (\frac{\pi }{4}-x+x)}{\cos (\frac{\pi }{4}-x)\cos x}dx$

$I=\frac{\pi }{8}\underset{0}{\overset{\pi /4}{\int }}\frac{\sin (\frac{\pi }{4}-x)\cos x+\cos (\frac{\pi }{4}-x)\sin x}{\cos (\frac{\pi }{4}-x)\cos x}dx$

$I=\frac{\pi }{8}\underset{0}{\overset{\pi /4}{\int }}[\tan (\frac{\pi }{4}-x)+\tan x]dx$

$\because \underset{0}{\overset{\pi /4}{\int }}\tan (\frac{\pi }{4}-x)dx=\underset{0}{\overset{\pi /4}{\int }}\tan xdx$

$\therefore I=\frac{\pi }{4}\underset{0}{\overset{\pi /4}{\int }}\tan xdx$

$\Rightarrow I=\frac{\pi }{4}[\ln |\sec x|{]}_0^{\pi /4}$

$\Rightarrow I=\frac{\pi }{4}\ln \sqrt{2}$

$\Rightarrow I=\frac{\pi }{8}\ln 2$


Now, when $m=\sqrt{2}$
$I^{\prime }=\underset{0}{\overset{\pi /2}{\int }}\frac{1}{1+(\tan x{)}^{\sqrt{2}}}dx...\left(3\right)$

Using $\underset{0}{\overset{a}{\int }}f\left(x\right)dx=\underset{0}{\overset{a}{\int }}f(a-x)dx$

$I^{\prime }=\underset{0}{\overset{\pi /2}{\int }}\frac{1}{1+(\cot x{)}^{\sqrt{2}}}dx$

$I^{\prime }=\underset{0}{\overset{\pi /2}{\int }}\frac{(\tan x{)}^{\sqrt{2}}}{1+(\tan x{)}^{\sqrt{2}}}dx...\left(4\right)$

Adding eqn $\left(3\right)$ and $\left(4\right)$, we get

$2I^{\prime }=\underset{0}{\overset{\pi /2}{\int }}1dx$

$\Rightarrow I^{\prime }=\frac{\pi }{4}$