The integral int_{0}^{frac{pi}{2}} frac{1}{3+ | vedclass

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Question

(B) Let $I = \int_{0}^{\frac{\pi}{2}} \frac{dx}{3+2 \sin x + \cos x}$.Using the substitution $	an(\frac{x}{2}) = t$,we have $dx = \frac{2 dt}{1+t^2}$

Vedclass · Accepted Answer

$	an^{-1}(2)-\frac{\pi}{4}$

Vedclass · Answer

(B) Let $I = \int_{0}^{\frac{\pi}{2}} \frac{dx}{3+2 \sin x + \cos x}$.Using the substitution $	an(\frac{x}{2}) = t$,we have $dx = \frac{2 dt}{1+t^2}$,$\sin x = \frac{2t}{1+t^2}$,and $\cos x = \frac{1-t^2}{1+t^2}$.Substituting these into the integral:$I = \int_{0}^{1} \frac{1}{3 + 2(\frac{2t}{1+t^2}) + \frac{1-t^2}{1+t^2}} \cdot \frac{2 dt}{1+t^2}$$I = \int_{0}^{1} \frac{2 dt}{3(1+t^2) + 4t + 1 - t^2} = \int_{0}^{1} \frac{2 dt}{2t^2 + 4t + 4} = \int_{0}^{1} \frac{dt}{t^2 + 2t + 2}$$I = \int_{0}^{1} \frac{dt}{(t+1)^2 + 1}$$I = [	an^{-1}(t+1)]_{0}^{1} = 	an^{-1}(2) - 	an^{-1}(1) = 	an^{-1}(2) - \frac{\pi}{4}$.

The integral $\int_{0}^{\frac{\pi}{2}} \frac{1}{3+2 \sin x+\cos x} d x$ is equal to.

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