MCQ
The value of $\int_0^{\pi / 2} \frac{d x}{1+\tan ^3 x}$ is
  • A
    $0$
  • B
    1
  • C
    $\frac{\pi}{2}$
  • $\frac{\pi}{4}$

Answer

Correct option: D.
$\frac{\pi}{4}$
(D)
Let $I=\int_0^{\frac{\pi}{2}} \frac{d x}{1+\tan ^3 x}$
$=\int_0^{\frac{\pi}{2}} \frac{\cos ^3 x}{\sin ^3 x+\cos ^3 x} d x$ ...(i)
$\therefore \quad I=\int_0^{\frac{\pi}{2}} \frac{\sin ^3 x}{\cos ^3 x+\sin ^3 x} d x$ ...(ii)
$\ldots\left[\because \int_0^{ a } f (x) d x=\int_0^{ a } f ( a -x) d x\right]$
Adding (i) and (ii), we get
$2 I =\int_0^{\frac{\pi}{2}} d x=[x]_0^{\pi / 2}$
$\therefore \quad 2 I =\frac{\pi}{2} \Rightarrow I =\frac{\pi}{4}$
Alternate Method:
$\int_0^{\frac{\pi}{2}} \frac{d x}{1+\tan ^{ n } x}=\int_0^{\frac{\pi}{2}} \frac{d x}{1+\cot ^{ n } x} d x=\frac{\pi}{4}$

$\int_0^{\pi / 2} \frac{d x}{1+\tan ^3 x}=\frac{\pi}{4}$ 

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