Question

\[\int\frac{\cos x - \sin x}{\sqrt{8 - \sin2x}}dx\]

Answer 1

\[\text{ Let I } = \int\frac{\cos x - \sin x}{\sqrt{8 - \sin2x}}dx\]
\[ = \int\frac{\cos x - \sin x}{\sqrt{9 - 1 - \sin2x}}dx\]
\[ = \int\frac{\cos x - \sin x}{\sqrt{9 - \sin^2 x - \cos^2 x - 2\sin x\cos x}}dx\]
\[ = \int\frac{\cos x - \sin x}{\sqrt{9 - \left( \sin x + \cos x \right)^2}}dx\]
\[ Let \left( \sin x + \cos x \right) = t\]
\[ \text{On differentiating both sides, we get}\]
\[ \left( \cos x - \sin x \right)dx = dt\]
\[ \therefore I = \int\frac{1}{\sqrt{\left( 3 \right)^2 - \left( t \right)^2}}dt\]
\[ = \sin^{- 1} \left( \frac{t}{3} \right) + c\]
\[ = \sin^{- 1} \left( \frac{\sin x + \cos x}{3} \right) + c\]
\[Hence, \int\frac{\cos x - \sin x}{\sqrt{8 - \ sin2x}}dx = \sin^{- 1} \left( \frac{\sin x + \cos x}{3} \right) + c\]