If int frac{2 cos x+3 sin x}{4 cos x+5 sin x} | vedclass

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Question

(B) Let $I = \int \frac{2 \cos x+3 \sin x}{4 \cos x+5 \sin x} dx$. We express the numerator as: $2 \cos x + 3 \sin x = A(4 \cos x + 5 \sin x) + B \fra

Vedclass · Accepted Answer

$\frac{-2}{41}$

Vedclass · Answer

(B) Let $I = \int \frac{2 \cos x+3 \sin x}{4 \cos x+5 \sin x} dx$. We express the numerator as: $2 \cos x + 3 \sin x = A(4 \cos x + 5 \sin x) + B \frac{d}{dx}(4 \cos x + 5 \sin x)$. $2 \cos x + 3 \sin x = A(4 \cos x + 5 \sin x) + B(-4 \sin x + 5 \cos x)$. $2 \cos x + 3 \sin x = (4A + 5B) \cos x + (5A - 4B) \sin x$. Comparing coefficients: $4A + 5B = 2$ and $5A - 4B = 3$. Multiplying the first by $4$ and second by $5$: $16A + 20B = 8$ and $25A - 20B = 15$. Adding them: $41A = 23 \implies A = \frac{23}{41}$. Substituting $A$ in $4A + 5B = 2$: $4(\frac{23}{41}) + 5B = 2 \implies \frac{92}{41} + 5B = 2 \implies 5B = 2 - \frac{92}{41} = \frac{82-92}{41} = \frac{-10}{41} \implies B = \frac{-2}{41}$. Thus,$\int \frac{2 \cos x+3 \sin x}{4 \cos x+5 \sin x} dx = \int \left( \frac{23}{41} + \frac{-2}{41} \frac{-4 \sin x + 5 \cos x}{4 \cos x + 5 \sin x} \right) dx$. $= \frac{23}{41} x - \frac{2}{41} \log |4 \cos x + 5 \sin x| + c$. Comparing with the given form,$K = \frac{-2}{41}$.

If $\int \frac{2 \cos x+3 \sin x}{4 \cos x+5 \sin x} dx = \left(\frac{23}{41}\right) x + K \log |4 \cos x+5 \sin x| + c$,then $K$ is equal to

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