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In evaluating integral $\int\sin^{3}x dx $ I am pretty sure we need to use substitution $e^x=t$, but can't go next step.

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    $\begingroup$ Why would you substitute $e^x$ when you have nothing like it in the integral? Wouldn't it help to rather substitute some trigonometric term? $\endgroup$
    – Arpan
    Mar 24, 2015 at 12:22
  • $\begingroup$ possible duplicate of Evaluating $\int P(\sin x, \cos x) \text{d}x$ $\endgroup$ Mar 24, 2015 at 12:42

4 Answers 4

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$$\sin^3 x \,dx = \sin^2 x \sin x = (1 - \cos^2 x) \sin x$$

Let $u = \cos x\implies du = -\sin x$.

That gives us: $$\int \sin^3 x \,dx = -\int (1 -u^2)\,du = \int (u^2 - 1)\,du$$

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Use this trignometric identity

$$\sin 3x=3\sin x-4\sin^3 x$$

$$\therefore \sin^3 x=\frac{3\sin x - \sin 3x}{4}$$

I think you can proceed from here.

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$$ \int\sin^{3}x~dx =\int \left( \dfrac{3\sin x-\sin 3x }{4}\right) ~dx =\dfrac{3}{4}\int \sin x~dx-\dfrac{1}{12}\int \sin 3x~d(3x)\\=-\dfrac{3\cos x}{4}+\dfrac{\cos 3x}{12}+C $$

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I don't know about the substitution you suggested, but you might have success if you use $\sin^2(x) + \cos^2(x) = 1$ so that $\sin^2(x) = 1 - \cos^2(x)$. Then this changes your problem:

$$\int \sin^3(x) \, dx = \int \sin(x) \, dx - \int \cos^2(x) \sin(x) \, dx$$

These two new integrals should be easier.

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