When doing this problem, we can see that the cosine is to an odd power. So we're going to pull out one of those cosines and then we're going to change everything else to sines. So this cosine squared X that's left is going to turn into one minus sine squared X DX. So we'd have three integral cosine X sine 4X DX -3 integral cosine X sine to the 6th power DX. Now if we think about letting U equals sine X then our DU is cosine X DX. So this turns into three three integral cosine XDX so U to the 4th DU -3 that cosine XDX is DU again, so integral U to the 6th DU. So now we get 3/5 U to the 5th -3 sevenths U to the 7th plus C. So if we substitute back in, we get 3/5 sine to the 5th X -, 3 sevenths sine to the seven X + C. Now if this is not an answer, that is one of our options. We could do things like switch our sines into cosines by using our Pythagorean identity.