Question

Solve :${\int }_0^{\pi /2}\frac{dx}{1+{\tan }^{3}x}$ ?

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

Answer 1

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

${\int }_0^{\frac{\pi }{2}}\frac{dx}{1+{\tan }^{3}x}={\int }_0^{\frac{\pi }{2}}\frac{1}{(1+u^2)(1+u^3)}du[let,u=\tan x]={\int }_0^{\frac{\pi }{2}}\frac{1-2u}{3(u^2-u+1)}+\frac{u+1}{2(u^2+1)}+\frac{1}{6(u+1)}du={\int }_0^{\frac{\pi }{2}}\frac{1-2u}{3(u^2-u+1)}du+{\int }_0^{\frac{\pi }{2}}\frac{u+1}{2(u^2+1)}du+{\int }_0^{\frac{\pi }{2}}\frac{1}{6(u+1)}du={[-\frac{1}{3}log[u^2-u+1]]}_0^{\frac{\pi }{2}}+{\left[\frac{1}{2}(\frac{1}{2}log(u^2+1)+{\tan }^{-1}\left(u\right))\right]}_0^{\frac{\pi }{2}}+{\left[\frac{1}{6}log(u+1)\right]}_{\left(0\right)}^{\frac{\pi }{2}}=\frac{\pi }{4}-0=\frac{\pi }{4}$