JEE Mains · Maths · STD 12 - 7.2 definite integral
The integral \(\int_{0}^{\frac{\pi}{2}} \frac{1}{3+2 \sin x+\cos x} d x\) is equal to.
- A \(\tan ^{-1}(2)\)
- B \(\tan ^{-1}(2)-\frac{\pi}{4}\)
- C \(\frac{1}{2} \tan ^{-1}(2)-\frac{\pi}{8}\)
- D \(\frac{1}{2}\)
Answer & Solution
Correct Answer
(B) \(\tan ^{-1}(2)-\frac{\pi}{4}\)
Step-by-step Solution
Detailed explanation
\(I=\int_{0}^{\frac{\pi}{2}} \frac{d x}{3+2 \sin x+\cos x}=\int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2} \frac{x}{2} \cdot d x}{2 \tan ^{2} \frac{x}{2}+4 \tan \frac{x}{2}+4}\) Put \(\tan \frac{x}{2}=t\), so…
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