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

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(B) We have, $\int \frac{a \cos x-2 \sin x}{b \sin x+5 \cos x} d x=\frac{7}{41} x+\frac{22}{41} \log |b \sin x+5 \cos x|+C$. Let $a \cos x-2 \sin x =

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$\frac{2}{\sqrt{7}} 	an ^{-1}\left(\frac{	an \frac{x}{2}}{\sqrt{7}}ight)+C$

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(B) We have, $\int \frac{a \cos x-2 \sin x}{b \sin x+5 \cos x} d x=\frac{7}{41} x+\frac{22}{41} \log |b \sin x+5 \cos x|+C$. Let $a \cos x-2 \sin x = \lambda(b \sin x+5 \cos x) + \mu(b \cos x-5 \sin x)$. Equating coefficients of $\sin x$ and $\cos x$: $a = 5\lambda + \mu b$ and $-2 = b\lambda - 5\mu$. Given $\lambda = \frac{7}{41}$ and $\mu = \frac{22}{41}$, we have: $a = 5(\frac{7}{41}) + b(\frac{22}{41}) = \frac{35+22b}{41}$ and $-2 = b(\frac{7}{41}) - 5(\frac{22}{41}) \Rightarrow -82 = 7b - 110 \Rightarrow 7b = 28 \Rightarrow b = 4$. Substituting $b=4$ into $a$: $a = \frac{35+22(4)}{41} = \frac{35+88}{41} = \frac{123}{41} = 3$. Now, $\int \frac{d x}{b+a \cos x} = \int \frac{d x}{4+3 \cos x}$. Using the substitution $	an \frac{x}{2} = t$, $\cos x = \frac{1-t^2}{1+t^2}$ and $dx = \frac{2 dt}{1+t^2}$: $\int \frac{2 dt}{(1+t^2)(4+3(\frac{1-t^2}{1+t^2}))} = \int \frac{2 dt}{4+4t^2+3-3t^2} = \int \frac{2 dt}{7+t^2}$. $= \frac{2}{\sqrt{7}} 	an^{-1}(\frac{t}{\sqrt{7}}) + C = \frac{2}{\sqrt{7}} 	an^{-1}\left(\frac{	an \frac{x}{2}}{\sqrt{7}}ight) + C$.

If $\int \frac{a \cos x-2 \sin x}{b \sin x+5 \cos x} d x=\frac{7}{41} x+\frac{22}{41} \log |b \sin x+5 \cos x|+C, (a>0, b>0)$, then $\int \frac{d x}{b+a \cos x}=$

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