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

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(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 $\cos x$:$1 = 4A - 6B$ and $8 = 6A + 4B$.Solving these equations: Multiply the first by $2$ and the second by $3$:$2 = 8A - 12B$ and $24 = 18A + 12B$.Adding them gives $26 = 26A$,so $A = 1$.Substituting $A = 1$ into $1 = 4(1) - 6B$,we get $6B = 3$,so $B = \frac{1}{2}$.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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