Mohammad nabeel's question at Yahoo Answers regarding a definite integral

[MarkFL]

Here is the original question:

Ntegral of sin^3(6x)cos^4(3x) dx ? between limits o to 60 degrees

Here is a link to the question:

Ntegral of sin^3(6x)cos^4(3x) dx ? - Yahoo! Answers

I have posted a link there to this topic so that the OP can find my response.

 

[MarkFL]

Hello Mohammad,

We are given to calculate:

$\displaystyle \int_0^{\frac{\pi}{3}}\sin^3(6x)\cos^4(3x)\,dx$

If we use a change of variables, i.e.,

$\displaystyle u=x-\frac{\pi}{6}\,\therefore\,du=dx\,\therefore\,x=u+\frac{\pi}{6}$

then we may rewrite the integrand as follows:

$\displaystyle \sin(6x)=\sin\left(6\left(u+\frac{\pi}{6} \right) \right)=\sin(6u+\pi)=\sin(6u)\cos(\pi)+\cos(6u) \sin(\pi)=-\sin(6u)$

$\displaystyle \cos(3x)=\cos\left(3\left(u+\frac{\pi}{6} \right) \right)=\cos\left(3u+\frac{\pi}{2} \right)=\cos(3u)\cos\left(\frac{\pi}{2} \right)-\sin(3u)\sin\left(\frac{\pi}{2} \right)=-\sin(3u)$

and our integral becomes (don't forget to change the limits in accordance with the change of variable):

$\displaystyle -\int_{-\frac{\pi}{6}}^{\frac{\pi}{6}}\sin^3(6u)\sin^4(3u)\,dx$

Now we have an odd-function as the integrand, and by the odd function rule, this is simply zero.