Integrating a sine function that is to an odd power

Question

I've started the chapter in my book where we begin to integrate trig functions, so bear in mind I've only got started and that I do not have a handle on more advanced techniques.

$\eqalign{ & \int {{{\sin }^3}x} dx \cr & = \int {\sin x({{\sin }^2}x} )dx \cr & = \int {\sin x({1 \over 2}} - {1 \over 2}\cos 2x)dx \cr & = \int {{1 \over 2}\sin x(1 - \cos 2x)dx} \cr & = \int {{1 \over 2}\sin x(1 - (2{{\cos }^2}x - 1))dx} \cr & = \int {{1 \over 2}\sin x(2 - 2{{\cos }^2}x)dx} \cr & = \int {\sin x - {{\cos }^2}} x\sin xdx \cr & y = {1 \over 3}{\cos ^3}x - \cos x + C \cr} $

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I got the right answer but it seems like an awfully long winded way of doing things, have I made things harder than they should be with this method?

Answer 1

You have unnecessary middle steps, just use $\sin^2x = 1-\cos^2x$ after the second line.

To handle other odd powers:

$$\sin^{2k+1}x=(\sin^{2}x)^k(\sin x)=(1-\cos^2x)^k(\sin x)$$

then use $u=\cos x$.

Answer 2

Isn't it faster to use subsitution like $\cos(x)=t$? $$\eqalign{ & \int {{{\sin }^3}x} dx \cr & = \int {\sin x({{\sin }^2}x} )dx \cr & =-\int \sin^2(x)d\cos(x) \cr & = -\int (1-\cos^2(x)) d\cos(x) \cr & = -\int (1-t^2) dt \cr } $$

Answer 3

Using the formula $$\sin (3x)=3\sin x-4\sin^3 x$$ you can calculate the integral in just one step.