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BITSAT 2021
Exam
BITSAT
Year
2021
Subject
Mathematics
Chapter
Definite Integration

BITSAT 2021 · Mathematics · Definite Integration

Evaluate: ∫_ 0 ^ π (1)/(5+4 x) d x

Evaluate: ∫0π15+4cos⁡xdx\int_{0}^{\pi} \frac{1}{5+4 \cos x} d x

Options

  1. Aπ3\frac{\pi}{3}
  2. Bπ2\frac{\pi}{2}
  3. Cπ4\frac{\pi}{4}
  4. Dπ6\frac{\pi}{6}

Correct answer

A. π3\frac{\pi}{3}

Step-by-step solution

We have, ∫0π15+4cos⁡xdx=∫0π15+4(1−tan⁡2x21+tan⁡2x2)dx=∫0π1+tan⁡2x25(1+tan⁡2x2)+4(1−tan⁡2x2)dx=∫0π1+tan⁡2x29+tan⁡2x2dx=∫0πsec⁡2x29+tan⁡2x2dx\int_{0}^{\pi} \frac{1}{5+4 \cos x} d x=\int_{0}^{\pi} \frac{1}{5+4\left(\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}\right)} d x=\int_{0}^{\pi} \frac{1+\tan ^{2} \frac{x}{2}}{5\left(1+\tan ^{2} \frac{x}{2}\right)+4\left(1-\tan ^{2} \frac{x}{2}\right)} d x=\int_{0}^{\pi} \frac{1+\tan ^{2} \frac{x}{2}}{9+\tan ^{2} \frac{x}{2}} \mathrm{dx}=\int_{0}^{\pi} \frac{\sec ^{2} \frac{x}{2}}{9+\tan ^{2} \frac{x}{2}} \mathrm{dx} Let tan⁡x2=t⇒12sec⁡2x2dx=dt\tan \frac{x}{2}=t \Rightarrow \frac{1}{2} \sec ^{2} \frac{x}{2} d x=d t Also, x=0⇒t=0x=0 \Rightarrow t=0 and x=π⇒t=∞∴x=\pi \Rightarrow t=\infty\therefore\mathrm{I}=∫0∞dt9+t2∴=\int_{0}^{\infty} \frac{\mathrm{dt}}{9+t^{2}}\therefore\mathrm{I}=2∫0∞dt9+t2∴=2 \int_{0}^{\infty} \frac{\mathrm{dt}}{9+t^{2}}\therefore\mathrm{I}=23=\frac{2}{3}\left[\tan ^{-1} \frac{t}{3}\right]_{0}^{\infty}=\frac{2}{3}[tan⁡−1∞−tan⁡−10]=23(π2−0)=π3\left[\tan ^{-1} \infty-\tan ^{-1} 0\right]=\frac{2}{3}\left(\frac{\pi}{2}-0\right)=\frac{\pi}{3}
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