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?

  • $\begingroup$ Your substitution is right. Now parse this integral into two integrals. In first make a substitution : $cosx = t$ and in second : $sin2x = 2sinxcosx$ $\endgroup$ – openspace Apr 20 '16 at 8:12
  • $\begingroup$ The calculator says that the integral does not converge. $\endgroup$ – Solumilkyu Apr 20 '16 at 8:55
  • $\begingroup$ Did you try $\sin(x)=u$ $\endgroup$ – Archis Welankar Apr 20 '16 at 9:34

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





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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.