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STD 12 - 7.2 definite integral — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 7.2 definite integral4 Marks
MCQ
The integral $\int_{0}^{\frac{\pi}{2}} \frac{1}{3+2 \sin x+\cos x} d x$ is equal to.
  • A
    $\tan ^{-1}(2)$
  • ✓
    $\tan ^{-1}(2)-\frac{\pi}{4}$
  • C
    $\frac{1}{2} \tan ^{-1}(2)-\frac{\pi}{8}$
  • D
    $\frac{1}{2}$
✓

Answer

Correct option: B.
$\tan ^{-1}(2)-\frac{\pi}{4}$
b
$I=\int_{0}^{\frac{\pi}{2}} \frac{d x}{3+2 \sin x+\cos x}=\int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2} \frac{x}{2} \cdot d x}{2 \tan ^{2} \frac{x}{2}+4 \tan \frac{x}{2}+4}$

Put $\tan \frac{x}{2}=t$, so

$I=\int_{0}^{1} \frac{d t}{(t+1)^{2}+1}=\left.\tan ^{-1}(x+1)\right|_{0} ^{1}=\tan ^{-1} 2-\frac{\pi}{4}$

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