Need a little help with integral of [sin2x]^3 * [cos2x]^2 dx

[Opus89]

I have a trig integral to do and I'm a little stuck. It is the integral of: [sin2x]^3 * [cos2x]^2 dx. The cosine being squared is what is throwing me off. If it wasn't squared, I could let u = sin2x and my du would = 2cos2x dx. So I would have u^3 du Which would be easy. Also, if I didn't have the 2x( just x by itself instead) to deal with in each term, it would also be easy. But, thats not the case and I'm not sure how to attack this problem. Any tips?

Thanks

 

[galactus]

Re: Need a little help with trigonometric integral

You could let \(\displaystyle cos^{2}(2x)=1-sin^{2}(2x)\) and rewrite as:

\(\displaystyle \int sin^{3}(2x)(1-sin^{2}(2x))dx=\int sin^{3}(2x)dx-\int sin^{5}(2x)dx\)

Or you could break off a factor of sin and rewrite as:

\(\displaystyle (1-cos^{2}(2x))sin(2x)cos^{2}(2x)\)

Let \(\displaystyle u=cos(2x), \;\ \frac{-du}{2}=sin(2x)dx\)

Rewrite:

\(\displaystyle \frac{-1}{2}\int (1-u^{2})u^{2}du=\frac{1}{2}\int u^{4}du-\frac{1}{2}\int u^{2}du\)

 

[Opus89]

Re: Need a little help with trigonometric integral

Ok, I did the integral and I'm hoping someone can tell me whether or not my answer is correct. I got, (cos[2x])^3 / 6 - (cos[2x])^5 / 10 + c. How does it look?

 

[Deleted member 4993]

Re: Need a little help with trigonometric integral

Ok, I did the integral and I'm hoping someone can tell me whether or not my answer is correct. I got, (cos[2x])^3 / 6 - (cos[2x])^5 / 10 + c. How does it look?

Differentiate the answer you got - then try to reduce that to the "integrant". If you can - you are most probably correct.

 

[Opus89]

Re: Need a little help with trigonometric integral

Ok, I did the integral and I'm hoping someone can tell me whether or not my answer is correct. I got, (cos[2x])^3 / 6 - (cos[2x])^5 / 10 + c. How does it look?

Differentiate the answer you got - then try to reduce that to the "integrant". If you can - you are most probably correct.

Oh, so I need to differentiate the answer I got and if that answer is the same as what I was trying to integrate then it's correct, right?