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

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(B) Let $I = \int \frac{a \cos x+3 \sin x}{5 \cos x+2 \sin x} d x$. We express the numerator as $A(	ext{denominator}) + B(\frac{d}{dx}(	ext{denomina

Vedclass · Accepted Answer

$11$

Vedclass · Answer

(B) Let $I = \int \frac{a \cos x+3 \sin x}{5 \cos x+2 \sin x} d x$. We express the numerator as $A(	ext{denominator}) + B(\frac{d}{dx}(	ext{denominator}))$. $a \cos x + 3 \sin x = A(5 \cos x + 2 \sin x) + B(-5 \sin x + 2 \cos x)$. Equating coefficients of $\cos x$ and $\sin x$: $5A + 2B = a$ $2A - 5B = 3$ Solving for $A$ and $B$: Multiply first by $5$ and second by $2$: $25A + 10B = 5a$ and $4A - 10B = 6$. Adding gives $29A = 5a + 6$,so $A = \frac{5a+6}{29}$. From $2A - 5B = 3$,$5B = 2A - 3 = 2(\frac{5a+6}{29}) - 3 = \frac{10a+12-87}{29} = \frac{10a-75}{29}$,so $B = \frac{10a-75}{145}$. Comparing with the given integral $\int (A + B \frac{-5 \sin x + 2 \cos x}{5 \cos x + 2 \sin x}) dx = Ax + B \log |5 \cos x + 2 \sin x| + c$. Given $A = \frac{26}{29}$,so $\frac{5a+6}{29} = \frac{26}{29} \implies 5a = 20 \implies a = 4$. Given $B = -\frac{k}{29}$,so $B = -\frac{k}{29} = \frac{10(4)-75}{145} = \frac{-35}{145} = -\frac{7}{29}$. Thus,$k = 7$. Finally,$|a+k| = |4+7| = 11$.

If $\int \frac{a \cos x+3 \sin x}{5 \cos x+2 \sin x} d x=\frac{26}{29} x-\frac{k}{29} \log |5 \cos x+2 \sin x|+c$,then $|a+k|=$

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