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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...

Thank you in advance!

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  • $\begingroup$ 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? ... $\endgroup$
    – mickep
    Mar 7, 2018 at 20:27
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    $\begingroup$ 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. $\endgroup$ Mar 7, 2018 at 20:28
  • $\begingroup$ 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... $\endgroup$ Mar 7, 2018 at 20:30

2 Answers 2

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HINT:

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 $-2\cdot\left(1-\text{s}^4\right)$

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  • $\begingroup$ Thank you, I will try that $\endgroup$ Mar 7, 2018 at 20:31
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Put $cosx=z^2$ then $$\int\frac{\sin^3{x}}{\sqrt{\cos{x}}}dx=\int2(z^4-1)dz.$$

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  • $\begingroup$ Thank you very much, that works perfectly $\endgroup$ Mar 7, 2018 at 22:21

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