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# Evaluate the following integrals:

Question:

Evaluate the following integrals:

$\int \frac{\cos x-\sin x}{\sqrt{8-\sin 2 x}} d x$

Solution:

$=\int \frac{\cos x-\sin x}{\sqrt{8-\sin 2 x}} d x=\int \frac{\sin x-\cos x}{\sqrt{8-(\sin x+\cos x)^{2}+1}} d x$

Let $\sin x+\cos x=t$

$(\cos x-\sin x)=d t$

Therefore, $\int \frac{\sin x-\cos x}{\sqrt{8-(\sin x+\cos x)^{2}+1}} d x=\int \frac{d t}{\sqrt{9-t^{2}}}$

Since we have, $\int \frac{1}{\sqrt{a^{2}-x^{2}}} d x=\sin ^{-1}\left(\frac{x}{a}\right)+c$

$=\int \frac{d t}{\sqrt{9-t^{2}}}=\int \frac{d t}{\sqrt{3^{2}-t^{2}}}=\sin ^{-1}\left(\frac{t}{3}\right)+c$

$=\sin ^{-1}\left(\frac{\sin x+\cos x}{3}\right)+c=\sin ^{-1}\left(\frac{\sin x}{3}+\frac{\cos x}{3}\right)+c=\sin ^{-1}\left(\frac{\sin x}{3}\right)+$

$\sin ^{-1}\left(\frac{\cos x}{3}\right)+c$

$=\frac{x}{3}+\sin ^{-1}\left(\frac{\sin x}{3}\right)+c$