# Integrating Powers of sin

Tags: integrating, powers
 P: 3,007 Okay so I am suposed to evaluate 2 of these: 1.) $$\int\sin^3(a\theta)d\theta$$ the solution manual looks like it used a trig ID to do this. Why?? Doesn't $$\int\sin^3u du=\frac{1}{3}\cos^3u-(\cos u)+C$$ ? So just use the u-sub $$u=a\theta \Rightarrow du/a=d\theta$$ So it should just be $$\frac{1}{3a}\cos^3a\theta-\frac{1}{a}\cosa\theta+C$$ ? But they got: $$-\frac{1}{a}\cos a\theta-\frac{1}{3a}\cos^3a\theta+C$$ Why the negative -1/3a ? What did I miss? Is it my formula? 2.) For $$\int\sin^4(3x)\cos^3xdx$$ they went from <---that to $$=\int\sin^4(3x)(1-sin^23x)\cos3xdx$$ How the hell does that make this problem ANY easier?! Is there a formula for ^^^that! Casey
 P: 440 You're using sin(u) like a variable, and integrating, which will certainly get you nowhere. One thing is that $$\int sin(u)^{3}$$ does not equal $$\frac{1}{3}\cos^3u$$. You're going to have to use u-substitution, as you already have, and almost certainly some integration by parts.
HW Helper
Thanks
P: 24,975

## Integrating Powers of sin

I would suggest you use the trig identity sin(x)^2=1-cos(x)^2, followed by u=cos(x).
 HW Helper P: 6,207 $$=\int\sin^4(3x)(1-sin^23x)\cos3xdx$$ If you let u=sin3x does that make it any easier?
P: 3,007
 Quote by Dick I would suggest you use the trig identity sin(x)^2=1-cos(x)^2, followed by u=cos(x).
I'll try that now. But why is my formula not producing the correct result if I am using u=a*theta ? Shouldn't it be 1/a *{Pug it in} ?
 P: 685 $$\int \sin^3(a\theta)d\theta = \int \sin(a\theta)d\theta - \int \sin(a\theta)\cos^2(a\theta)d\theta$$ The first term should be easy for you to integrate, for the second term, try a substitution of $$u = \cos(a\theta) \Rightarrow \frac{du}{d\theta} = -a\sin(a\theta) \Rightarrow \sin(a\theta)d\theta = -\frac{1}{a}du$$ After which the equation simplifies a little. Use a similar idea for the second one.
P: 3,007
 Quote by Dick I would suggest you use the trig identity sin(x)^2=1-cos(x)^2, followed by u=cos(x).
So I got it using this. But I am still trying to figure out where I went wrong with the formula.

Casey
HW Helper
P: 2,884
 Quote by Saladsamurai So I got it using this. But I am still trying to figure out where I went wrong with the formula. Casey
Seems to me like there is simply a typo in the book (the coefficient of the cos^3 should indeed be positive)
P: 3,007
 Quote by nrqed Seems to me like there is simply a typo in the book (the coefficient of the cos^3 should indeed be positive)
Not if you use the trig substitution; it is negative. It is definitely a mistake on my part.

Casey
 Sci Advisor HW Helper Thanks P: 24,975 No, it's positive. (1-cos^(x)^2)*sin(x)*dx. u=cos(x), du=-sin(x)*dx. The u^3/3 term comes out with a plus sign.
P: 1,757
 Quote by Saladsamurai 2.) For $$\int\sin^4(3x)\cos^3xdx$$ they went from <---that to $$=\int\sin^4(3x)(1-sin^23x)\cos3xdx$$
did you mean $$\cos^{3}3xdx$$?
P: 3,007
 Quote by rocophysics did you mean $$\cos^{3}3xdx$$?
I don't know what I meant?! I figured it out though.

Thanks!
Casey
 P: 1,757 i was trying to solve it for myself but then it saw that ... lol, is all good though.

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