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} $

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?

http://trevorpythag.co.uk/2009/mathematics/calculus/intergration-using-a-reduction-formula-with-example/ — lab bhattacharjee, May 14 '13 at 16:16
Related : http://math.stackexchange.com/questions/236543/proving-a-reduction-formula-for-the-antiderivative-of-cosnx — lab bhattacharjee, May 14 '13 at 16:18
You could have skipped a bunch of steps: $ \int \sin^3xdx
= \int \sin x(1-\cos^2x)dx $. Then let $u=1-\cos^2x$. — Brady Trainor, May 14 '13 at 16:18

vadim123 · Accepted Answer · 2013-05-14T16:30:52.240

7

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 $u=\cos x$.

edited May 14 '13 at 16:30

answered May 14 '13 at 16:18

vadim123

82,796

I see this now, thank you! – seeker May 14 '13 at 16:22
@Assad: this works for any odd power of $\sin x$ – Ross Millikan May 14 '13 at 16:23
@RossMillikan, Could you explain why just odd powers please? – seeker May 14 '13 at 16:26
1

@Assad Think to $\int\sin^2 x,dx$, which isn't a polynomial in $\sin x$. – egreg May 14 '13 at 16:31
3

@Assad: because you convert an even number of powers of $\sin x$ to $(1-cos^2 x)$, leaving a $\sin x dx $ over to allow the $u$ substitution $u=\cos x, du=-\sin x dx$. If it is an even power of sine that doesn't work. There are other ways, a bit harder. – Ross Millikan May 14 '13 at 16:31

score 2 · Answer 2 · answered May 14 '13 at 16:27

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

score 2 · Answer 3 · answered May 14 '13 at 16:29

2

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

answered May 14 '13 at 16:29

pritam

10,157

Integrating a sine function that is to an odd power

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