MHT CET · Maths · Indefinite Integration
\(\int_0^{\frac{\pi}{2}} \frac{d x}{5+4 \cos x}=\)
- A \(\frac{1}{3} \tan ^{-1}\left(\frac{1}{3}\right)\)
- B \(2 \tan ^{-1}\left(\frac{1}{3}\right)\)
- C \(\frac{2}{3} \tan ^{-1}\left(\frac{1}{3}\right)\)
- D \(\tan ^{-1}\left(\frac{1}{3}\right)\)
Answer & Solution
Correct Answer
(C) \(\frac{2}{3} \tan ^{-1}\left(\frac{1}{3}\right)\)
Step-by-step Solution
Detailed explanation
\(\int_0^{\frac{\pi}{2}} \frac{d x}{5+4 \cos x}=\int_0^{\frac{\pi}{2}} \frac{d x}{5+4 \cdot \frac{1-\tan ^2 \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}}=\) \(\int_0^{\frac{\pi}{2}} \frac{\sec ^2 \frac{x}{2} d x}{9+\tan ^2 \frac{x}{2}}=2 \int_0^{\frac{\pi}{2}} \frac{\frac{1}{2} \sec ^2 \frac{x}{2}}{3^2+\tan ^2 \frac{x}{2}}\)
\(x=2 \int_0^1 \frac{d t}{3^2+t^2} \frac{1}{3}=2 \times\left[\tan ^{-1} \frac{t}{3}\right]_0^1=2\)
\(=\frac{2}{3}\left\{\tan ^{-1} \frac{1}{3}-\tan ^{-1} 0\right\}=\frac{2}{3} \tan ^{-1} \frac{1}{3}\)
\(x=2 \int_0^1 \frac{d t}{3^2+t^2} \frac{1}{3}=2 \times\left[\tan ^{-1} \frac{t}{3}\right]_0^1=2\)
\(=\frac{2}{3}\left\{\tan ^{-1} \frac{1}{3}-\tan ^{-1} 0\right\}=\frac{2}{3} \tan ^{-1} \frac{1}{3}\)
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