Question

The integral $\int\limits_0^{\frac{\pi}{2}} \frac{1}{3+2 \sin x+\cos x} d x$ 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 1

Answer: (B)

$I=\int\limits_0^{\frac{\pi}{2}} \frac{d x}{3+2 \sin x+\cos x}=\int\limits_0^{\frac{\pi}{2}} \frac{\sec ^2 \frac{x}{2} \cdot d x}{2 \tan ^2 \frac{x}{2}+4 \tan \frac{x}{2}+4}$
Put $\tan \frac{x}{2}=t$, so
$I=\int\limits_0^1 \frac{d t}{(t+1)^2+1}=\left.\tan ^{-1}(x+1)\right|_0 ^1=\tan ^{-1} 2-\frac{\pi}{4}$