Question

${\int }_0^{\pi /2}\frac{dx}{1+{\tan }^{3}x}$ is equal to

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

Answer 1

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

Let $l={\int }_0^{\pi /2}\frac{dx}{1+{\tan }^{3}x}$
$\Rightarrow l={\int }_0^{\pi /2}\frac{{\cos }^{3}x}{{\sin }^{3}x+{\cos }^{3}x}dx$ ... (i)
$\Rightarrow l={\int }_0^{\pi /2}\frac{{\cos }^3(\frac{\pi }{2}-x)}{{\sin }^3(\frac{\pi }{2}-x)+{\cos }^3(\frac{\pi }{2}-x)}dx\{\because {\int }_0^{a}f(x)={\int }_0^{a}f(a-x\left)dx\right\}$
$\Rightarrow l={\int }_0^{\pi /2}\frac{{\sin }^{3}x}{{\cos }^{3}x+{\sin }^{3}x}dx$ ....(ii)
On adding Eqs. (i) and (ii), we get
$2l={\int }_0^{\pi /2}\frac{{\sin }^{3}x+{\cos }^{3}x}{{\sin }^{3}x+{\cos }^{3}x}dx$
$={\int }_0^{\pi /2}\left(1\right)dx=[x{]}_0^{\pi /2}$
$\Rightarrow 2l=\frac{\pi }{2}\Rightarrow l=\frac{\pi }{4}$.