1got it simple it was.i had not tried it.[4][4]
\int \frac{cosx - sinx}{\sqrt{8-sin2x}}dx
ans-----> sin ^{-1}\left(\frac{sinx + cosx}{3} \right)+c
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2 Answers
1write sin2x as
( sinx + cosx ) ^2 = 1 + sin2x
sin2x = ( sinx + cosx ) ^2 - 1
\int \frac{cosx - sinx }{\sqrt{8 - [ ( sinx + cosx ) ^2 -1 )] }}
\int \frac{cosx - sinx }{\sqrt{9 - ( sinx + cosx ) ^2 }}
put sinx + cosx = t
cosx - sinx dx = dt
\int \frac{dt }{\sqrt{9 - ( t ) ^2 }} \rightarrow \sin^{-1}\left( \frac{t}{3}\right)