# How can I solve$\int \sin^3(x)dx$?

I have to find the integral $$\int \sin^3xdx\\= \int \sin^2x \sin xdx\\= \int (1-\cos^2x) \sin xdx$$

Substitution: $$z=\cos x$$ $$\frac{dz}{dx} = -\sin x$$ $$-dz = \sin x dx$$

Now the above expression would be like this $$\int -(1-z^2) dz$$ Now integration would be $$-z + \frac{z^3}{3} + c$$ we replace $z$ by $\cos x$ so our answer would be $$-\cos x + \frac{\cos^3x}{3} + c$$

But in book this answer is not correct. I want to know the error. Please, can any one solve it and tell me about the error?

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Please format your question using $\LaTeX$ enclosed in dollar signs. – Your Ad Here Jan 27 '14 at 10:43
I'm sorry to ask you such a personal question zoonie, but how old are you? – Tomáš Zato Jan 27 '14 at 10:48
@zonnie, it isn't nice to ask a further question in the comments without even addressing other comments... – DonAntonio Jan 27 '14 at 10:50
Your answer is correct; there is no error. Nice solution, in fact. The answer in the book may have simplified $-\cos x +(\cos^3 x)/3$ further. – David Mitra Jan 27 '14 at 11:14
@zonnie. May be, you could answer the questions you have been asked. – Claude Leibovici Jan 27 '14 at 11:30

This method looks easier.You can use $sin3x$=3$sinx$-4$sin^3x$. Hence you will get $sin^3x$=$\frac{3sinx-sin3x}{4}$.Hence $$\int \sin^3(x)dx\\=\int \frac{3sinx-sin3x}{4}= \int \frac{3sinx}{4}-\int \frac{sin3x}{4}=-\frac{3cosx}{4}+\frac{cos3x}{12}+c$$
Why $\sin$ instead of $\cos$? – apnorton Jan 27 '14 at 12:55