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

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(D) Let $I = \int \frac{3 \cos x-2 \sin x}{4 \sin x+5 \cos x} d x = A \log |5 \cos x+4 \sin x|+B x+c$. By differentiating both sides with respect to $

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$A=\frac{22}{41}$ and $B=\frac{7}{41}$

Vedclass · Answer

(D) Let $I = \int \frac{3 \cos x-2 \sin x}{4 \sin x+5 \cos x} d x = A \log |5 \cos x+4 \sin x|+B x+c$. By differentiating both sides with respect to $x$,we get: $\frac{3 \cos x-2 \sin x}{4 \sin x+5 \cos x} = A \frac{d}{dx}(\log |5 \cos x+4 \sin x|) + B$ $\frac{3 \cos x-2 \sin x}{4 \sin x+5 \cos x} = A \frac{-5 \sin x+4 \cos x}{5 \cos x+4 \sin x} + B$ $\frac{3 \cos x-2 \sin x}{4 \sin x+5 \cos x} = \frac{A(-5 \sin x+4 \cos x) + B(4 \sin x+5 \cos x)}{4 \sin x+5 \cos x}$ Equating the numerators: $3 \cos x-2 \sin x = (4A+5B) \cos x + (-5A+4B) \sin x$ Comparing coefficients of $\cos x$ and $\sin x$: $4A+5B = 3$  --- $(1)$ $-5A+4B = -2$ --- $(2)$ Multiplying $(1)$ by $5$ and $(2)$ by $4$: $20A+25B = 15$ $-20A+16B = -8$ Adding these equations: $41B = 7 \Rightarrow B = \frac{7}{41}$. Substituting $B$ in $(1)$: $4A + 5(\frac{7}{41}) = 3 \Rightarrow 4A = 3 - \frac{35}{41} = \frac{123-35}{41} = \frac{88}{41} \Rightarrow A = \frac{22}{41}$. Thus,$A = \frac{22}{41}$ and $B = \frac{7}{41}$.

If $\int \frac{3 \cos x-2 \sin x}{4 \sin x+5 \cos x} d x=A \log |5 \cos x+4 \sin x|+B x+c$,then $A$ and $B$ are

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