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Evaluate the following integrals:

Question:

Evaluate the following integrals:

$\int \frac{\cos 2 x}{\sqrt{\sin ^{2} 2 x+8}} d x$

Solution:

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

$=\frac{1}{2} \int \mathrm{dt} / \sqrt{\left(\mathrm{t}^{2}+(2 \sqrt{2})^{2}\right.}=\frac{1}{2} \log \left[\mathrm{t}+\sqrt{\mathrm{t}^{2}+8}\right]+\mathrm{c}$

$=\frac{1}{2} \log \left[\mathrm{t}+\sqrt{\mathrm{t}^{2}+8}\right]+\mathrm{c}=\frac{1}{2} \log \left[\sin 2 \mathrm{x}+\sqrt{\sin ^{2} 2 \mathrm{x}+8}\right]+\mathrm{c}$

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