Q89 / 180 · COMEDK 2012 · Mathematics

$\int_{0}^{\pi / 2} \frac{\sin 2 t}{\sin ^{4} t+\cos ^{4} t} d t=$

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

Solution

We have, $$ \begin{aligned} I &=\int_{0}^{\pi / 2} \frac{\sin 2 t}{\sin ^{4} t+\cos ^{4} t} d t \\ &=\int_{0}^{\pi / 2} \frac{2 \sin t \cos t}{\sin ^{4} t+\cos ^{4} t} d t \\ &=\int_{0}^{\pi / 2} \frac{2 \tan t \sec ^{2} t}{\left(\tan ^{2} t\right)^{2}+1} d t \end{aligned} $$ Put $\tan ^{2} t=x$ $$ \begin{aligned} &\left(2 \tan t \sec ^{2} t\right) d t=d x \\ &\therefore \quad I=\int_{0}^{\infty} \frac{1}{1+x^{2}} d x=\left[\tan ^{-1} x\right]_{0}^{\infty}=\frac{\pi}{2} \end{aligned} $$

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

Relevant Questions