int frac{sin x+8 cos x}{4 sin x+6 cos x} d x | vedclass

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Question

(A) Let $I = \int \frac{\sin x+8 \cos x}{4 \sin x+6 \cos x} d x$.We express the numerator as $A(	ext{denominator}) + B(\frac{d}{dx}(	ext{denominator

Vedclass · Accepted Answer

$x+\frac{1}{2} \log |4 \sin x+6 \cos x|+c$

Vedclass · Answer

(A) Let $I = \int \frac{\sin x+8 \cos x}{4 \sin x+6 \cos x} d x$.We express the numerator as $A(	ext{denominator}) + B(\frac{d}{dx}(	ext{denominator}))$:$\sin x+8 \cos x = A(4 \sin x+6 \cos x) + B(4 \cos x-6 \sin x)$.Equating the coefficients of $\sin x$ and $\cos x$:For $\sin x$: $4A - 6B = 1$.For $\cos x$: $6A + 4B = 8$.Solving these equations,we multiply the first by $2$ and the second by $3$:$8A - 12B = 2$ and $18A + 12B = 24$.Adding them gives $26A = 26$,so $A = 1$.Substituting $A=1$ into $4A - 6B = 1$ gives $4 - 6B = 1$,so $6B = 3$,$B = \frac{1}{2}$.Thus,$I = \int \frac{(4 \sin x+6 \cos x) + \frac{1}{2}(4 \cos x-6 \sin x)}{4 \sin x+6 \cos x} d x$.$I = \int 1 d x + \frac{1}{2} \int \frac{4 \cos x-6 \sin x}{4 \sin x+6 \cos x} d x$.$I = x + \frac{1}{2} \log |4 \sin x+6 \cos x| + c$.

$\int \frac{\sin x+8 \cos x}{4 \sin x+6 \cos x} d x$ is equal to

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