Q116 / 160 · AP EAMCET 2006 · Mathematics

$\int_0^{\pi / 2} \frac{d x}{1+\tan ^3 x}$ is equal to :

  • A
    $\pi$
  • B
    $\frac{\pi}{2}$
  • Correct answer
    C
    $\frac{\pi}{4}$
  • D
    $\frac{3 \pi}{2}$

Solution

$I=\int_0^{\pi / 2} \frac{d x}{1+\tan ^3 x}$ $I=\int_0^{\pi / 2} \frac{\cos ^3 x}{\sin ^3 x+\cos ^3 x} d x$ ...(i) $I=\int_0^{\pi / 2} \frac{\cos ^3\left(\frac{\pi}{2}-x\right)}{\sin ^3\left(\frac{\pi}{2}-x\right)+\cos ^3\left(\frac{\pi}{2}-x\right)} d x$ $I=\int_0^{\pi / 2} \frac{\sin ^3 x}{\sin ^3 x+\cos ^3 x} d x$ ...(ii) On adding Eqs. (i) and (ii), we get $2 I=\int_0^{\pi / 2} \frac{\sin ^3 x+\cos ^3 x}{\sin ^3 x+\cos ^3 x} d x$ $=\int_0^{\pi / 2} 1 d x=\frac{\pi}{2}$ $\Rightarrow I=\frac{\pi}{4}$

Hence, the correct answer is option: 'C'.

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