Q74 / 125 · BITSAT 2013 · Mathematics

Elvaluate $\int_{0}^{\pi / 2} \frac{\sin x}{1+\cos ^{2} x} d x$

  • A
    $\pi / 2$
  • Correct answer
    B
    $\pi / 4$
  • C
    $\pi / 3$
  • D
    $\pi$

Solution

Let $\mathrm{I}=\int_{0}^{\pi / 2} \frac{\sin \mathrm{x}}{1+\cos ^{2} \mathrm{x}} \mathrm{dx}$ Let $\cos \mathrm{x}=\mathrm{t}$ and $-\sin \mathrm{x} \mathrm{d} \mathrm{x}=\mathrm{dt}$ Now, $\mathrm{x}=0 \Rightarrow \mathrm{t}=\cos 0=1$ and $\mathrm{x}=\frac{\pi}{2} \Rightarrow \mathrm{t}=\cos \frac{\pi}{2}=0$ $\therefore \quad \mathrm{I}=\int_{1}^{0} \frac{\sin \mathrm{x}}{1+\mathrm{t}^{2}}\left(\frac{-\mathrm{dt}}{\sin \mathrm{x}}\right)=-\int_{1}^{0} \frac{\mathrm{dt}}{1+\mathrm{t}^{2}}$ $=\left[\tan ^{-1} \mathrm{t}\right]_{1}^{0}=-\left[0-\frac{\pi}{4}\right]=\frac{\pi}{4}$

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

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