Question

Assertion :STATEMENT-1 : ${\int }_0^{\pi /2}\frac{dx}{1+{\tan }^{3}x}=\frac{\pi }{4}$ Reason: STATEMENT-2 : ${\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f(x+a)dx$ where $f\left(x\right)$ is an integrable function.

• A. Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1
• B. Statement-1 is True, Statement-2 is True; Statement-2 is Not a correct explanation for Statement-1
• C. Statement-1 is True, Statement-2 is False
• D. Statement-1 is False, Statement-2 is True

Answer 1

The correct option is C Statement-1 is True, Statement-2 is False

Let $I={\int }_0^{\frac{\pi }{4}}\frac{1}{1+{\tan }^{3}x}dx$
Substitute $t=\tan x\Rightarrow dt={\sec }^{2}xdx$
$\therefore I={\int }_0^1\frac{1}{(t^2+1)(t^3+3)}dt$
$={\int }_0^1(\frac{1-2t}{3(t^2-t+1)}+\frac{t+1}{2(t^2+1)}+\frac{1}{6(t+1)})dt$
$={\int }_0^1(\frac{1-2t}{3(t^2-t+1)}+\frac{t}{t^2+1}+\frac{1}{t^2+1}+\frac{1}{6(t+1)})dt$
$={[\frac{1}{4}\log (t^2+1)-\frac{1}{3}\log (t^2-t+1)+\frac{1}{6}\log (t+1)+\frac{1}{2}{\tan }^{-1}t+c]}_0^1=\frac{\pi }{4}$
Reason is false as
${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$