Question

Value of ${\int }_0^{\pi /4}(\pi x-4x^2)\log (1+\tan x)dx$ is

• A. $\frac{{\pi }^3}{192}\log 2$
• B. $\frac{{\pi }^3}{192}\log \sqrt{2}$
• C. $\frac{{\pi }^3}{36}\log 2$
• D. $\frac{{\pi }^3}{48}\log \sqrt{2}$

Answer 1

Answer: (A)

$\frac{{\pi }^3}{192}\log 2$

Let, $I={\int }_0^{\pi /4}(\pi x-4x^2)\log (1+\tan x)dx={\int }_0^{\pi /4}4x(\frac{\pi }{4}-x)\log (1+\tan \left(x\right))dx$
Using the property, ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$, we get
$I={\int }_0^{\pi /4}4(\frac{\pi }{4}-x)x\log (1+\tan (\frac{\pi }{4}-x))dx$
$\because \tan (\frac{\pi }{4}-x)=\frac{1-\tan x}{1+\tan x}$
$⟹I={\int }_0^{\pi /4}4x(\frac{\pi }{4}-x)\log \left(\frac{1+\tan x+1-\tan x}{1+\tan x}\right)dx$
$={\int }_0^{\pi /4}4x(\frac{\pi }{4}-x)\log \left(\frac{2}{1+\tan x}\right)dx$
$={\int }_0^{\pi /4}4x(\frac{\pi }{4}-x)\left[\log \left(2\right)-\log (1+\tan x)\right]dx$
$={\int }_0^{\pi /4}4x(\frac{\pi }{4}-x)\log \left(2\right)dx-{\int }_0^{\pi /4}4x(\frac{\pi }{4}-x)\log (1+\tan x)dx$
$⟹I={\int }_0^{\pi /4}4x(\frac{\pi }{4}-x)\log \left(2\right)dx-I$
$⟹2I={\int }_0^{\pi /4}4x(\frac{\pi }{4}-x)\log \left(2\right)dx$
$⟹I=\frac{\log 2}{2}{\int }_0^{\pi /4}(\pi x-4x^2)dx$
$=\frac{1}{2}{|\frac{\pi x^2}{2}-\frac{4x^3}{3}|}_0^{\pi /4}\log 2$
By applying limits, we get $I=\frac{{\pi }^3}{192}\log 2$

Hence, option A is correct.