make the substitution $\displaystyle u= sin(\theta) => du= cos(\theta) d\theta$
Now our integral becomes $\displaystyle \int u^6 du$ which is much easier to solve => solve using the general power rule
Can you finish it off ?
Ah okay now it is easy. if sin to the six theta is simply sin theta to the power of six, then you can substitute 'just the sin theta' . I didn't know if that six in there made it one whole thing but now I see that it's simple. I've never seen that before. Yes I've got it! Thank you both! (:
There is a problem after that (#27) that is almost exactly the same thing:
I would expect everything to be the same except it is now 5 theta. But they are saying you should get (1/35) times sin of 5 theta to the seven instead of what I got which was (1/7) times that. Confused...
this problem is a little trickier => make the substitution $\displaystyle u=5\theta => du=5d\theta$ initially
now our integral becomes $\displaystyle \frac{1}{5} \int sin^{6}(u) cos(u) du$
now make the substitution $\displaystyle v=sin(u) => dv=cos(u) du$
$\displaystyle = \frac{1}{5} \int v^{6} dv$ now solve this simply integral using the general power rule.