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

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Question

(C) Let $I = \int \frac{2 \sin x - 3 \cos x}{4 \cos x - 3 \sin x} dx$. We express the numerator as $A \cdot (	ext{denominator}) + B \cdot \frac{d}{dx

Vedclass · Accepted Answer

$\frac{1}{25}[\log |4 \cos x - 3 \sin x| - 18x] + c$

Vedclass · Answer

(C) Let $I = \int \frac{2 \sin x - 3 \cos x}{4 \cos x - 3 \sin x} dx$. We express the numerator as $A \cdot (	ext{denominator}) + B \cdot \frac{d}{dx}(	ext{denominator})$. $2 \sin x - 3 \cos x = A(4 \cos x - 3 \sin x) + B(-4 \sin x - 3 \cos x)$. Equating coefficients of $\sin x$ and $\cos x$: $-3A - 4B = 2$ and $4A - 3B = -3$. Solving these equations: Multiply the first by $3$ and the second by $4$: $-9A - 12B = 6$ and $16A - 12B = -12$. Subtracting: $25A = -18 \implies A = -\frac{18}{25}$. Substituting $A$: $4(-\frac{18}{25}) - 3B = -3 \implies -\frac{72}{25} + 3 = 3B \implies 3B = \frac{3}{25} \implies B = \frac{1}{25}$. Thus, $I = \int \frac{-\frac{18}{25}(4 \cos x - 3 \sin x) + \frac{1}{25}(-4 \sin x - 3 \cos x)}{4 \cos x - 3 \sin x} dx$. $I = -\frac{18}{25} \int 1 dx + \frac{1}{25} \int \frac{-4 \sin x - 3 \cos x}{4 \cos x - 3 \sin x} dx$. $I = -\frac{18}{25}x + \frac{1}{25} \log |4 \cos x - 3 \sin x| + c$. Rearranging gives $\frac{1}{25}[\log |4 \cos x - 3 \sin x| - 18x] + c$.

$\int \frac{2 \sin x - 3 \cos x}{4 \cos x - 3 \sin x} dx = $

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