int frac{3cos x + 2sin x}{4sin x + 5cos x} dx | vedclass

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Question

(A) Let $I = \int \frac{3\cos x + 2\sin x}{4\sin x + 5\cos x} dx$. We express the numerator as $A(	ext{denominator}) + B(	ext{derivative of denomina

Vedclass · Accepted Answer

$1/41, \log |x|$

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(A) Let $I = \int \frac{3\cos x + 2\sin x}{4\sin x + 5\cos x} dx$. We express the numerator as $A(	ext{denominator}) + B(	ext{derivative of denominator})$. Let $3\cos x + 2\sin x = \lambda(4\sin x + 5\cos x) + \mu(4\cos x - 5\sin x)$. Comparing coefficients of $\sin x$ and $\cos x$: $2 = 4\lambda - 5\mu$ and $3 = 5\lambda + 4\mu$. Solving these equations: Multiply the first by $4$ and second by $5$: $8 = 16\lambda - 20\mu$ and $15 = 25\lambda + 20\mu$. Adding them gives $23 = 41\lambda$,so $\lambda = 23/41$. Substituting $\lambda$ into $2 = 4(23/41) - 5\mu$: $5\mu = 92/41 - 82/41 = 10/41$,so $\mu = 2/41$. Thus,$I = \int \frac{\lambda(4\sin x + 5\cos x) + \mu(4\cos x - 5\sin x)}{4\sin x + 5\cos x} dx = \lambda x + \mu \ln |4\sin x + 5\cos x| + c$. $I = \frac{23}{41}x + \frac{2}{41} \ln |4\sin x + 5\cos x| + c = \frac{1}{41} \{23x + 2 \ln |4\sin x + 5\cos x|\} + c$. Comparing with the given form,$A = 1/41$ and $f(x) = \ln |x|$.

$\int \frac{3\cos x + 2\sin x}{4\sin x + 5\cos x} dx = A \{23x + 2\ln |4\sin x + 5\cos x|\} + c$,then $A$ and $f(x)$ are

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