Evaluate the following integral:
$$\int \frac{\sqrt{\sin ^4x+\cos ^4x}}{\sin ^3x \cos x }dx$$ where $x \in \big(0,\frac{\pi}{2} \big)$
Could some give me hint as how to approach this question?
I tried to use the fact that $\sin ^4x+\cos ^4x=1-\frac{\sin^22x}{2}$ but it didn't help. How should I proceed?
On
Dividing by $\cos^2x$ on the top and bottom gives $\displaystyle\int\frac{\sqrt{\tan^4x+1}}{\frac{\sin^3x}{\cos x}}dx=\int\frac{\sqrt{\tan^4x+1}}{\tan^3x}\sec^2x \;dx$.
Now let $u=\tan x$ to get $\displaystyle\int\frac{\sqrt{u^4+1}}{u^3}du,\;$ and then let $t=u^2$ to get $\displaystyle\frac{1}{2}\int\frac{\sqrt{t^2+1}}{t^2}dt$.
Next let $t=\tan\theta\;$ to get
$\displaystyle\frac{1}{2}\int\frac{\sec\theta}{\tan^2\theta}\;\sec^2\theta\;d\theta=\frac{1}{2}\int\frac{\sec\theta}{\tan^2\theta}\;(\tan^2\theta+1)\;d\theta=\frac{1}{2}\int\big(\sec\theta+\cot\theta\csc\theta\big)d\theta$
$\displaystyle=\frac{1}{2}\big[\ln\big|\sec\theta+\tan\theta\big|-\csc\theta\big]+C=\frac{1}{2}\left[\ln\left(\sqrt{1+u^4}+u^2\right)-\frac{\sqrt{1+u^4}}{u^2}\right]+C$
$\displaystyle=\frac{1}{2}\left[\ln\left(\sqrt{1+\tan^4x}+\tan^2x\right)-\frac{\sqrt{1+\tan^4x}}{\tan^2x}\right]+C$
Using your substitution: $$\int{\frac{dx}{sin^{3}x\cdot cosx}-2\int{\frac{cosx dx}{sinx}}}$$
Now using substitution : $t = sinx$: $$\int{\frac{dx}{sin^{3}x\cdot cosx} = \int{\frac{du}{u^{3}(1-u^{2})}}}$$, you could finish it by parsing your integral by some parts. And of course second integral you should know.