If int frac{(cos x-sin x)}{8-sin 2 x} d x=fra | vedclass

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(C) Let $I = \int \frac{\cos x - \sin x}{8 - \sin 2x} dx$. We know that $\sin 2x = 1 - (1 - \sin 2x) = 1 - (\sin x - \cos x)^2$. However,it is easier

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(C) Let $I = \int \frac{\cos x - \sin x}{8 - \sin 2x} dx$. We know that $\sin 2x = 1 - (1 - \sin 2x) = 1 - (\sin x - \cos x)^2$. However,it is easier to write $8 - \sin 2x = 9 - (1 + \sin 2x) = 9 - (\sin x + \cos x)^2$. So,$I = \int \frac{\cos x - \sin x}{3^2 - (\sin x + \cos x)^2} dx$. 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 standard 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 $p = 6$.

If $\int \frac{(\cos x-\sin x)}{8-\sin 2 x} d x=\frac{1}{p} \log \left[\frac{3+\sin x+\cos x}{3-\sin x-\cos x}\right]+c$,then $p=$ (Where $c$ is a constant of integration)

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