Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have an integral here that I'm trying to figure out.

$$ \int 7\sin^2x \cos^4x\ dx $$

Here's what I got as an answer:

$$ \frac{7}{16}x-\frac{7}{64}\sin4x+\frac{7}{12}\sin2x + C $$

However, I'm doubting myself and the check didn't seem to produce good results. I can give some steps if you want. I filled a whole page with work, and it seems like it should be easier than that. Any ideas?

share|improve this question
    
wolframalpha.com/input/?i=int+7sin%28x%29%5E2+cos%28x%29%5E4 Click on "show steps" for one possible derivation. –  Bill Cook Oct 10 '11 at 0:41
    
@BillCook: That uses the reduction formula though. I'd rather stick to identities and such. –  confused Oct 10 '11 at 0:43
add comment

2 Answers

up vote 4 down vote accepted

You can use $\sin(x)\cos(x) = {1 \over 2}\sin(2x)$ and $\cos^2(x) = {1 + \cos(2x) \over 2}$ and your integral is $$7 \int\left({1 \over 2}\sin(2x)\right)^2 {1 + \cos(2x) \over 2}\,dx$$ $$= {7 \over 8}\int \sin^2(2x)\,dx + {7 \over 8}\int\sin^2(2x)\cos(2x)\,dx$$ The first term is just a $\sin^2$ integral, and the second can be dealt with by a $u$ substitution $u = \sin^2(2x)$.

share|improve this answer
    
That looks easy. I actually did the first step that you suggested, but somehow I made it a lot harder than it needed to be. Thanks for the help! –  confused Oct 10 '11 at 1:35
    
Hmm, I did it again when I had the chance and I got this: $$ \frac{7}{16}x-\frac{7}{64}\sin4x+\frac{7}{12}\sin^32x + C $$ Unfortunately, that doesn't seem to check according to WolframAlpha. –  confused Oct 10 '11 at 14:34
    
@confused I think the coefficient before the $\sin^3(2x)$ term should be ${7 \over 48}$. In other words your $du$ should be $2 \cos(2x)\,dx$ and not $2 du = \cos(2x)\,dx$. –  Zarrax Oct 10 '11 at 21:44
    
Yep, I just found that out. I had taken the antiderivative of $u$ instead of the derivative. However, even with that coefficient it still doesn't check. –  confused Oct 10 '11 at 21:49
    
There are multiple ways of writing a given function sometimes, so most likely they are the same function but in different forms. –  Zarrax Oct 10 '11 at 22:05
add comment

Hint: Try writing $\sin^2(x)=1-\cos^2(x)$ and then use the identity $\cos^2(x)=\frac{1+\cos(2x)}{2}$.

share|improve this answer
3  
...some algebra...apply identities again...some algebra...pain and suffering. –  Bill Cook Oct 10 '11 at 0:50
    
I'll give that a shot. I find it hard to know where to start on these things. –  confused Oct 10 '11 at 0:54
1  
"pain and suffering" - my take was "algebra, sweat, and tears"... –  J. M. Oct 10 '11 at 1:22
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.