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? ... $\endgroup$– mickepMar 7, 2018 at 20:27
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1$\begingroup$ 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. $\endgroup$– John HughesMar 7, 2018 at 20:28
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$\begingroup$ 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... $\endgroup$– Luka DuranovicMar 7, 2018 at 20:30
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2 Answers
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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 $-2\cdot\left(1-\text{s}^4\right)$
answered Mar 7, 2018 at 20:28
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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