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BITSAT 2013
Exam
BITSAT
Year
2013
Subject
Mathematics
Chapter
Definite Integration

BITSAT 2013 · Mathematics · Definite Integration

Elvaluate ∫_ 0 ^ π / 2 x 1+ ^ 2 x d x

Elvaluate ∫0π/2sin⁡x1+cos⁡2xdx\int_{0}^{\pi / 2} \frac{\sin x}{1+\cos ^{2} x} d x

Options

  1. Aπ/2\pi / 2
  2. Bπ/4\pi / 4
  3. Cπ/3\pi / 3
  4. Dπ\pi

Correct answer

B. π/4\pi / 4

Step-by-step solution

Let I=∫0π/2sin⁡x1+cos⁡2xdx\mathrm{I}=\int_{0}^{\pi / 2} \frac{\sin \mathrm{x}}{1+\cos ^{2} \mathrm{x}} \mathrm{dx} Let cos⁡x=t\cos \mathrm{x}=\mathrm{t} and −sin⁡xdx=dt-\sin \mathrm{x} \mathrm{d} \mathrm{x}=\mathrm{dt} Now, x=0⇒t=cos⁡0=1\mathrm{x}=0 \Rightarrow \mathrm{t}=\cos 0=1 and x=π2⇒t=cos⁡π2=0∴\mathrm{x}=\frac{\pi}{2} \Rightarrow \mathrm{t}=\cos \frac{\pi}{2}=0\therefore \quad\mathrm{I}=∫10sin⁡x1+\mathrmt2=\int_{1}^{0} \frac{\sin \mathrm{x}}{1+\mathrmt^{2}}\left(\frac{-\mathrm{dt}}{\sin \mathrm{x}}\right)=-\int_{1}^{0} \frac{\mathrm{dt}}{1+\mathrmt^{2}}=[tan⁡−1t]10=−[0−π4]=π4\left[\tan ^{-1} \mathrm{t}\right]_{1}^{0}=-\left[0-\frac{\pi}{4}\right]=\frac{\pi}{4}
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