int frac{cos x-sin x}{8-sin 2x} dx = frac{1}{ | vedclass

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(A) Let $I = \int \frac{\cos x-\sin x}{8-\sin 2x} dx$. We know that $\sin 2x = 2 \sin x \cos x$ and $1 = \sin^2 x + \cos^2 x$. So,$8 - \sin 2x = 9 - (

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$6$

Vedclass · Answer

(A) Let $I = \int \frac{\cos x-\sin x}{8-\sin 2x} dx$. We know that $\sin 2x = 2 \sin x \cos x$ and $1 = \sin^2 x + \cos^2 x$. So,$8 - \sin 2x = 9 - (1 + 2 \sin x \cos x) = 9 - (\sin x + \cos x)^2$. Thus,$I = \int \frac{\cos x - \sin x}{3^2 - (\sin x + \cos x)^2} dx$. Let $t = \sin x + \cos x$. Then $dt = (\cos x - \sin x) dx$. Substituting these into the integral,we get $I = \int \frac{dt}{3^2 - t^2}$. Using the formula $\int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \log \left| \frac{a+x}{a-x} \right| + C$,we have: $I = \frac{1}{2(3)} \log \left| \frac{3+t}{3-t} \right| + C = \frac{1}{6} \log \left| \frac{3+\sin x + \cos x}{3 - (\sin x + \cos x)} \right| + C$. Comparing this with the given expression $\frac{1}{p} \log \left[\frac{3+\sin x+\cos x}{3-\sin x-\cos x}\right]+c$,we find $p = 6$.

$\int \frac{\cos x-\sin x}{8-\sin 2x} dx = \frac{1}{p} \log \left[\frac{3+\sin x+\cos x}{3-\sin x-\cos x}\right]+c$,then $p = \ldots$

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