Question

Find the value of ${\int }_0^{\pi /2}\frac{dx}{1+{\tan }^{3}x}$

• A. $0$
• B. $1$
• C. $\frac{\pi }{2}$
• D. $\frac{\pi }{4}$

Answer 1

The correct option is D $\frac{\pi }{4}$

Let $I={\int }_0^{\pi /2}\frac{dx}{1+{\tan }^{3}x}$ ....(i)

As ${\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f(a-x)dx$
Therefore, $I={\int }_0^{\pi /2}\frac{dx}{1+{\tan }^3(\frac{\pi }{2}-x)}={\int }_0^{\pi /2}\frac{dx}{1+{\cot }^{3}x}$
$\Rightarrow I={\int }_0^{\pi /2}\frac{{\tan }^{3}x}{1+{\tan }^{3}x}dx$ ...(ii)
On adding equations (i) and (ii), we get
$2I={\int }_0^{\pi /2}\left(\frac{1+{\tan }^{3}x}{1+{\tan }^{3}x}\right)dx={\int }_0^{\pi /2}dx={\left[x\right]}_0^{\pi /2}$

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