Integral question I would like to ask for help with $$\int \cos^3 (x/9) \,dx$$
I got $$9\sin(x/9) - 3\sin^3(x/9) + c$$ but the book gives another answer which I'm not sure how they got to. 
 A: Observe that 
$$cos^3(y)=cos(y)cos^2(y)=\frac{1}{2}cos(y)(1+cos(2y))=\frac{1}{2}(cos(y)+cos(y)cos(2y))$$ 
In addition,
$$cos(y)cos(2y)=\frac{1}{2}(cos(y)+cos(3y))$$
Thus, 
$$\int cos^3(x/9)dx=9\int cos^3(x/9)d(x/9)=9\int cos^3(y)dy=9\int \frac{1}{2}(cos(y)+\frac{1}{2}(cos(y)+cos(3y)))dy=\frac{3}{4}\int cos(y)dy+\frac{1}{4}\int cos(3y)dy=...$$
I think you can finish it by yourself.
A: 
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I agree with you...
A: First use the substitution
$=\int \cos ^3\left(u\right)9du$
$=\int \:9\cos ^3\left(u\right)du$
Now take out the Constant
$=9\int \cos ^3\left(u\right)du$
Now use the reduction rule:
$\int \cos ^n\left(x\right)dx=\frac{\sin \left(x\right)\cos ^{n-1}\left(x\right)}{n}+\frac{n-1}{n}\int \cos ^{n-2}\left(x\right)dx$
$\int \cos ^3\left(u\right)du=\frac{\sin \left(u\right)\cos ^2\left(u\right)}{3}+\frac{2\int \cos \left(u\right)du}{3}$
$=9\left(\frac{\sin \left(u\right)\cos ^2\left(u\right)}{3}+\frac{2\int \cos \left(u\right)du}{3}\right)$
Substitute Back u
$=9\left(\frac{\sin \left(\frac{x}{9}\right)\cos ^2\left(\frac{x}{9}\right)}{3}+\frac{2\sin \left(\frac{x}{9}\right)}{3}\right)$
$=\frac{3\sin \left(\frac{x}{9}\right)\left(\cos \left(\frac{2x}{9}\right)+5\right)}{2}$
Now add the Constant
$=\frac{3\sin \left(\frac{x}{9}\right)\left(\cos \left(\frac{2x}{9}\right)+5\right)}{2}+C$
Hope this Helps!
