# How to integrate this using u substitution?

I have another rather simple problem that I cannot seem to be able to solve. I cannot find the right substitution.

The problem is:

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

I would appreciate any help...

• OK, so you have two trigonometric functions in your expression. Did you think of something? Did you try something? Perhaps $u=\sin x$? What happened? Maybe $u=\cos x$? What now? ... Mar 7, 2018 at 20:27
• You can show us your partial work by clicking "edit" just below your question; that'll let us know (a) that you've made some effort, and (b) how much you already understand. Mar 7, 2018 at 20:28
• I really meant to do so, but I haven't made any tangible progress. I tried with $u=\cos{x}$ but it didn't help. I cannot solve that third power of the sine function - it always remains and creates a problem... Mar 7, 2018 at 20:30

Use a substitution $\text{u}=\cos\left(x\right)$ then the integrand changes to $-\frac{1-\text{u}^2}{\sqrt{\text{u}}}$ then substitute $\text{s}:=\sqrt{\text{u}}$ then the integrand changes to $-2\cdot\left(1-\text{s}^4\right)$
Put $cosx=z^2$ then $$\int\frac{\sin^3{x}}{\sqrt{\cos{x}}}dx=\int2(z^4-1)dz.$$