Question

The value of $\int \frac{\sin x+\cos x}{\sqrt{1-\sin 2x}}dx$ is equal to
(a) $\sqrt{\sin 2x}+C$
(b) $\sqrt{\cos 2x}+C$
(c) ± (sin x − cos x) + C
(d) ± log (sin x − cos x) + C

Answer 1

(d) ± log (sin x − cos x) + C

$$\begin{gathered} \mathrm{Let}I=\int \frac{\left(\sin x+\cos x\right)dx}{\sqrt{1-\sin 2x}} \\ =\int \frac{\left(\sin x+\cos x\right)dx}{\sqrt{{\sin }^{2}x+{\cos }^{2}x-2\sin x\cos x}} \\ =\int \frac{\left(\sin x+\cos x\right)dx}{\sqrt{{\left(\sin x-\cos x\right)}^2}} \\ =\int \frac{\left(\sin x+\cos x\right)dx}{\left|\sin x-\cos x\right|} \\ =\pm \int \left(\frac{\sin x+\cos x}{\sin x-\cos x}\right)dx \\ \mathrm{Let}\sin x-\cos x=t \\ \Rightarrow \left(\cos x+\sin x\right)dx=dt \\ \therefore I=\pm \int \frac{dt}{t} \\ =\pm \ln \left|t\right|+\mathrm{C} \\ =\pm \ln \left|\sin x-\cos x\right|+C\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\left(\mathit{\because }\mathit{}t\mathit{=}\sin x\mathit{-}\cos \mathit{}x\right) \end{gathered}$$