Evaluate the following integrals:
$\int \frac{\cos 2 x}{\sqrt{\sin ^{2} 2 x+8}} d x$
Let $=\sin 2 x$
$d t=2 \cos 2 x d x$
$\cos 2 x d x=d t / 2$
$\int \frac{\cos 2 \mathrm{x}}{\sqrt{\sin ^{2} 2 \mathrm{x}+8}} \mathrm{dx}=\frac{1}{2} \int \mathrm{dt} / \sqrt{\left(\mathrm{t}^{2}+(2 \sqrt{2})^{2}\right.}$
Since we have, $\int \frac{1}{\sqrt{\left(x^{2}+a^{2}\right)}} d x=\log \left[x+\sqrt{\left.\left(x^{2}+a^{2}\right)\right]+c}\right.$
$=\frac{1}{2} \int \mathrm{dt} / \sqrt{\left(\mathrm{t}^{2}+(2 \sqrt{2})^{2}\right.}=\frac{1}{2} \log \left[\mathrm{t}+\sqrt{\mathrm{t}^{2}+8}\right]+\mathrm{c}$
$=\frac{1}{2} \log \left[\mathrm{t}+\sqrt{\mathrm{t}^{2}+8}\right]+\mathrm{c}=\frac{1}{2} \log \left[\sin 2 \mathrm{x}+\sqrt{\sin ^{2} 2 \mathrm{x}+8}\right]+\mathrm{c}$
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.