Deepak Scored 45->99%ile with Bounce Back Crack Course. You can do it too!

Evaluate the following integrals:


Evaluate the following integrals:

$\int \frac{\sin 8 x}{\sqrt{9+\sin ^{4} 4 x}} d x$


Let $t=\sin ^{2} 4 x$

$d t=2 \sin 4 x \cos 4 x \times 4 d x$

we know $\sin 2 x=2 \sin 2 x \cos 2 x$

therefore, $d t=4 \sin 8 x d x$

or, $\sin 8 x d x=d t / 4$

$\int \frac{\sin 8 x}{\sqrt{9+\sin ^{4} x}} d x=\frac{1}{4} \int \frac{d t}{\sqrt{3^{2}+t^{2}}}$

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

$=\frac{1}{4} \int \frac{d t}{\sqrt{3^{2}+t^{2}}}=\frac{1}{4} \log \left[t+\sqrt{t^{2}+3^{2}}+c\right.$

$=\frac{1}{4} \log \left[\sin ^{2} 4 x+\sqrt{9+\sin ^{4} 4 x}+c\right.$

Leave a comment

Free Study Material