int frac{13 cos 2 x-9 sin 2 x}{3 cos 2 x-4 si | vedclass

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Question

(A) Let $I = \int \frac{13 \cos 2 x-9 \sin 2 x}{3 \cos 2 x-4 \sin 2 x} d x$. We express the numerator as $A(	ext{denominator}) + B(\frac{d}{dx}(	ext

Vedclass · Accepted Answer

$3 x-\frac{1}{2} \log |3 \cos 2 x-4 \sin 2 x|+c$

Vedclass · Answer

(A) Let $I = \int \frac{13 \cos 2 x-9 \sin 2 x}{3 \cos 2 x-4 \sin 2 x} d x$. We express the numerator as $A(	ext{denominator}) + B(\frac{d}{dx}(	ext{denominator}))$. $13 \cos 2 x-9 \sin 2 x = A(3 \cos 2 x-4 \sin 2 x) + B(-6 \sin 2 x-8 \cos 2 x)$. Equating coefficients of $\cos 2 x$ and $\sin 2 x$: $3A - 8B = 13$ and $-4A - 6B = -9$. Solving these equations: Multiply first by $3$ and second by $4$: $9A - 24B = 39$ and $-16A - 24B = -36$. Subtracting gives $25A = 75$,so $A = 3$. Substituting $A=3$ into $3(3) - 8B = 13 \implies 9 - 8B = 13 \implies -8B = 4 \implies B = -\frac{1}{2}$. Thus,$I = \int \frac{3(3 \cos 2 x-4 \sin 2 x) - \frac{1}{2}(-6 \sin 2 x-8 \cos 2 x)}{3 \cos 2 x-4 \sin 2 x} d x$. $I = \int 3 d x - \frac{1}{2} \int \frac{-6 \sin 2 x-8 \cos 2 x}{3 \cos 2 x-4 \sin 2 x} d x$. $I = 3x - \frac{1}{2} \log |3 \cos 2 x-4 \sin 2 x| + c$.

$\int \frac{13 \cos 2 x-9 \sin 2 x}{3 \cos 2 x-4 \sin 2 x} d x=$

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