Question

Integrate

$\int \frac{\sqrt{{\sin }^{4}x+{\cos }^{4}x}}{{\sin }^{3}x\cos x}dx,$

Answer 1

$\int \frac{\sqrt{\sin {}^{4}x+\cos {}^{4}x}}{\sin {}^{3}x\cos x}dx=\int \frac{\sqrt{1+\tan {}^{4}x}}{\frac{\sin {}^{3}x}{\cos x}}dx=\int \frac{\sqrt{1+\tan {}^{4}x}}{\frac{\sin {}^{3}x}{\cos {}^{3}x}\cos {}^{2}x}dx=\int \frac{\sqrt{1+\tan {}^{4}x}}{\tan {}^{3}x}\sec {}^{2}xdx$

Now let $\tan x=u\sec {}^{2}xdx=du$

$=\int \frac{\sqrt{1+u^4}}{u^3}du$

Now let $u^2=t2u.du=dt\Rightarrow du=\frac{dt}{2\sqrt{t}}$

$=\int \frac{\sqrt{1+t^2}}{t^{\frac{3}{2}}}\frac{dt}{2\sqrt{t}}=\frac{1}{2}\int \frac{\sqrt{1+t^2}}{t^2}dt$

Now let $t=\tan \theta \Rightarrow dt=\sec {}^2\theta d\theta$

$=\frac{1}{2}\int \frac{\sqrt{1+\tan {}^2\theta }}{\tan {}^2\theta }\left(\sec {}^2\theta \right)d\theta =\frac{1}{2}\int \frac{\sec \theta \sec {}^2\theta }{\tan {}^2\theta }d\theta =\frac{1}{2}\int \sec \theta \left(\frac{1+\tan {}^2\theta }{\tan {}^2\theta }\right)d\theta =\frac{1}{2}\int \sec \theta (\cot {}^2\theta +1)d\theta =\frac{1}{2}\int \sec \theta .\cot {}^2\theta d\theta +\frac{1}{2}\int \sec \theta d\theta =\frac{1}{2}\int \sec \theta d\theta +\frac{1}{2}\int \cot \theta .\csc \theta d\theta =\frac{1}{2}\left(\ln |\sec \theta +\tan \theta |-\csc \theta \right)+C=\frac{1}{2}\left[\ln [\sqrt{1+u^4}+u^2]-\frac{\sqrt{1+u^4}}{u^2}\right]+C=\frac{1}{2}[\ln (\sqrt{1+\tan {}^{4}x}+\tan {}^{2}x)-\frac{\sqrt{1+\tan {}^{4}x}}{\tan {}^{2}x}]+C$