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$$\int \sin^3(x)dx$$ Using integration by parts: $$u=\sin^3(x)$$ $$u'=3\sin^2(x)\cos(x)$$ $$v'=1$$ $$v=x$$ Gives: $$\int \sin^3(x)dx= \sin^3(x)\cdot x - 3 \int \sin^2(x)\cdot cos(x)dx$$

Using: $$t=\sin(x)$$ $$\frac{dt}{dx}=\cos(x)$$ $$dt=\cos(x)dx$$

Results in: $$=\sin^3(x)\cdot x - \sin^3(x) +C$$

I used integration by parts method

Martin Sleziak
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4 Answers4

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By integration by parts you should have $$\int \sin^3(x)dx= \sin^3(x)\cdot x - 3 \int x\sin^2(x)\cdot \cos(x)dx.$$ The integral can be done in this way: $$\int \sin^3(x)\, dx=\int (1-\cos^2(x))\sin(x)\, dx=\int \sin(x) dx+\int \cos^2(x)D(\cos(x))\, dx.$$ Can you take it from here?

Robert Z
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  • Yes, thanks, I know it's a correct way to solve it but I'm trying to find where did I make a mistake? –  Jun 20 '17 at 14:02
  • @Natalia By integration by parts you should have $$\int \sin^3(x)dx= \sin^3(x)\cdot x - 3 \int x\sin^2(x)\cdot \cos(x)dx.$$ – Robert Z Jun 20 '17 at 14:09
  • @Natlia Is it clear now where you made the mistake? – Robert Z Jun 20 '17 at 14:17
  • I get it now thanks! What a stupid mistake I made –  Jun 20 '17 at 14:20
  • You forgot, Natalia, to use $uv- \int vu'$, in particular, you forgot to include v = x$ in the remaining integral. – amWhy Jun 20 '17 at 14:21
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If you use $u=\sin^3x$ and $v=1$, you get into big troubles, because the next integral you need to compute is $$ \int 3x\sin^2x\cos x\,dx $$ which is not at all easier than the one you started from.

If you really want to do it by parts, consider $u=\sin^2x$ and $v=\sin x$, so you get \begin{align} I&=\int\sin^3x\,dx \\ &=-\sin^2x\cos x+\int2\sin x\cos^2x\,dx \\ &=-\sin^2x\cos x+2\int\sin x\,dx-2\int\sin^3x\,dx \\ &=-\sin^2x\cos x-2\cos x-2I \end{align} that gives $$ I=\frac{1}{3}(-\sin^2x\cos x-2\cos x)+c=\frac{1}{3}(\cos^3x-3\cos x) $$

Much easier is to consider $$ \int\sin^3x\,dx=\int(1-\cos^2x)\sin x\,dx\underset{u=\cos x}{=}\int(u^2-1)\,du $$ which works alike for every odd power of $\sin x$.

egreg
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Use $$\sin^3x=\frac{3\sin{x}-\sin3x}{4},$$ but the Robert Z's and the Jack D'Aurizio's way is more nicer.

In your solution you need to write $x$ inside the last integral. This is your mistake.

Because $\int{v'}dx=\int1dx=x+C$.

Michael Rozenberg
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$$\int\sin^3x\,\mathrm{d}x=\int\sin^2x\,\mathrm{d}(\sin x)\ldots$$

using power-reducation formula you can continue without by parts

remember that $$\sin^2x=\frac{1-\cos2x}{2}$$

M.Mass
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