int_0^{pi / 2} frac{cos x}{3 cos x+sin x} d x | vedclass

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Question

(D) Let $I = \int_0^{\pi / 2} \frac{\cos x}{3 \cos x + \sin x} dx$. We express the numerator as: $\cos x = A(3 \cos x + \sin x) + B \frac{d}{dx}(3 \co

Vedclass · Accepted Answer

$\frac{3 \pi}{20}-\frac{\log 3}{10}$

Vedclass · Answer

(D) Let $I = \int_0^{\pi / 2} \frac{\cos x}{3 \cos x + \sin x} dx$. We express the numerator as: $\cos x = A(3 \cos x + \sin x) + B \frac{d}{dx}(3 \cos x + \sin x)$. $\cos x = A(3 \cos x + \sin x) + B(-3 \sin x + \cos x)$. Comparing coefficients of $\cos x$ and $\sin x$: $3A + B = 1$ and $A - 3B = 0$. From the second equation,$A = 3B$. Substituting into the first: $3(3B) + B = 1 \implies 10B = 1 \implies B = \frac{1}{10}$. Then $A = 3(\frac{1}{10}) = \frac{3}{10}$. Now,$I = \int_0^{\pi / 2} \left( \frac{3}{10} \frac{3 \cos x + \sin x}{3 \cos x + \sin x} + \frac{1}{10} \frac{-3 \sin x + \cos x}{3 \cos x + \sin x} \right) dx$. $I = \frac{3}{10} \int_0^{\pi / 2} dx + \frac{1}{10} \int_0^{\pi / 2} \frac{d(3 \cos x + \sin x)}{3 \cos x + \sin x}$. $I = \frac{3}{10} [x]_0^{\pi / 2} + \frac{1}{10} [\log |3 \cos x + \sin x|]_0^{\pi / 2}$. $I = \frac{3}{10} (\frac{\pi}{2} - 0) + \frac{1}{10} (\log |3(0) + 1| - \log |3(1) + 0|)$. $I = \frac{3 \pi}{20} + \frac{1}{10} (\log 1 - \log 3) = \frac{3 \pi}{20} - \frac{\log 3}{10}$.

$\int_0^{\pi / 2} \frac{\cos x}{3 \cos x+\sin x} d x=$

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