Question

$\int \frac{\sqrt{{\sin }^{4}x+{\cos }^{4}x}}{{\sin }^{3}x\cos x}dx,x\in \left(0,\frac{\pi }{2}\right)$

Answer 1

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

${({\sin }^{2}x+{\cos }^{2}x)}^2-2{\sin }^{2}x\times {\cos }^{2}x={sin}^{4}x+{\cos }^{4}x$

$(1-2{\sin }^{2}x{\cos }^{2}x)={\sin }^{4}x{\cos }^{4}x$

$\Rightarrow \int \frac{\sqrt{1-2{\sin }^{2}x{\cos }^{2}x}}{{\sin }^{3}x\cos x}dx$

Putting $\sin x=\frac{tanx}{\sec x}$

$\cos x=\frac{1}{\sec x}$

$\int \frac{\left({sec}^{2}x\right)({\tan }^{2}x+1)\sqrt{1-\frac{2{\tan }^{2}x}{{({\tan }^{2}x+1)}^2}}dx}{{\tan }^{3}x}$

Putting $\tan x=m$

${sec}^{2}xdx=dm$

$\int \frac{m^2+1\sqrt{1-\frac{{2m}^2}{{(m^2+1)}^2}}}{m^3}dm$

$\Rightarrow \int \frac{2m(m^2+1)\sqrt{1-\frac{{2m}^2}{{(m^2+1)}^2}}}{{2m}^4}dm$

Putting $m^2=t$

$2mdm=dt$

$\Rightarrow \int \frac{(t+1)\sqrt{1-\frac{2t}{{(t+1)}^2}}}{{2t}^2}dt$ Taking $(t+1)$ inside the root

$\Rightarrow \int \frac{\sqrt{{(t+1)}^2-2t}}{{2t}^2}dt=\int \frac{\sqrt{t^2+1}}{{2t}^2}dt$

Integrating by parts

$\Rightarrow f=\sqrt{t^2+1},g^1=\frac{1}{t^2}$

$\frac{df}{dt}=\frac{t}{\sqrt{t^2+1}};g=\frac{-1}{t}$

$=\frac{-\sqrt{t^2+1}}{2t}+\frac{1}{2}\int \frac{1}{\sqrt{t^2+1}}dt\to$ Standard integral

$=\frac{-\sqrt{t^2+1}}{2t}+\frac{1}{2}In(\sqrt{t^2+1}+t)+c$

$=\frac{1}{2}In|\sqrt{m^4+1}+m^2|-\frac{\sqrt{m^4+1}}{2m^2}+c=\frac{1}{2}In|\sqrt{{tan}^{4}x+1}+{tan}^{2}x|-\frac{\sqrt{{tan}^{4}x+1}}{2{tan}^{2}x}+c$