The value of $\int_0^{\frac{\pi}{2}} \frac{\mathrm{d} x}{1+\tan ^3 x}$

The value of $\int_0^{\frac{\pi}{2}} \frac{\mathrm{d} x}{1+\tan ^3 x}$
  1. 0
  2. $\frac{\pi}{4}$
  3. $\frac{\pi}{2}$
  4. 1

Solution

$\begin{aligned} & \int_0^{\frac{\pi}{2}} \frac{\mathrm{d} x}{1+\tan ^3 x}=\int_0^{\frac{\pi}{2}} \frac{\cos ^3 x}{\cos ^3 x+\sin ^3 x} \mathrm{~d} x=\frac{\frac{\pi}{2}-0}{2}=\frac{\pi}{4} \\ & {\left[\because \int_a^b \frac{f(x) \mathrm{d} x}{f(x)+f(a+b-x)}=\frac{b-a}{2}\right]}\end{aligned}$

Asked in: MHT CET 2022 (11 Aug Shift 1)

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