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

Question

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?

Answer 1

This method looks easier. You can use $\sin3x=3\sin x-4\sin^3x$. Hence you will get $\sin^3x=\frac{3\sin x-\sin(3x)}{4}$. Hence \begin{align}\int \sin^3(x)dx&=\int \frac{3\sin x-\sin(3x)}{4}\\&= \int \frac{3\sin x}{4}-\int \frac{\sin(3x)}{4}\\&=-\frac{3\cos x}{4}+\frac{\cos(3x)}{12}+c\end{align}