Question

The value of \[\int\frac{\sin x + \cos x}{\sqrt{1 - \sin 2x}} dx\] is equal to

Options

• \[\sqrt{\sin 2x} + C\]
• \[\sqrt{\cos 2x} + C\]

•  ± (sin x − cos x) + C

•  ± log (sin x − cos x) + C

Answer 1

± log (sin x − cos x) + C

 

\[\text{Let }I = \int\frac{\left( \sin x + \cos x \right) dx}{\sqrt{1 - \sin 2x}}\]

\[ = \int\frac{\left( \sin x + \cos x \right) dx}{\sqrt{\sin^2 x + \cos^2 x - 2 \sin x \cos x}}\]

\[ = \int\frac{\left( \sin x + \cos x \right) dx}{\sqrt{\left( \sin x - \cos x \right)^2}}\]

\[ = \int\frac{\left( \sin x + \cos x \right) dx}{\left| \sin x - \cos x \right|}\]

\[ = \pm \int\left( \frac{\sin x + \cos x}{\sin x - \cos x} \right)dx\]

\[\text{Let }\sin x - \cos x = t\]

\[ \Rightarrow \left( \cos x + \sin x \right)dx = dt\]

\[ \therefore I = \pm \int\frac{dt}{t}\]

\[ = \pm \ln \left| t \right| + C\]

\[ = \pm \ln \left| \sin x - \cos x \right| + C .............\left( \because t = \sin x - \cos x \right)\]