Evaluate the following integrals:


Evaluate the following integrals:

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


$\left.\int \frac{\sin x-\cos x}{\sqrt{\sin 2 x}} d x=\int(\sin x-\cos x) / \sqrt{(}(\sin x+\cos x)^{2}-1\right) d x$

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

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

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

Since, $\int \frac{1}{\sqrt{\left(x^{2}-a^{2}\right)}} d x=\log \left[x+\sqrt{\left(x^{2}-a^{2}\right)}\right]+c$

$=\int-\frac{d t}{\sqrt{t^{2}-1}}=-\log \left[t+\sqrt{t^{2}-1}\right]+c$

$=-\log \left[t+\sqrt{t^{2}-1}\right]+c=-\log [\sin x+\cos x+\sqrt{\sin 2 x}]+c$

Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now