How do I integrate 0.5sin^2(2x)cos(x) ? It's an AQA past question and I don't understand the mark scheme!
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- Thread Starter
- 12-05-2015 12:35
- 12-05-2015 12:46
What do you have now?
- 12-05-2015 21:15
(Original post by ladygabriela97)
- 12-05-2015 21:30
I did this question the other day! Yep, using the double angle formula for sin2x , squaring that all out, giving cos^3(x)sin^2(x) then replacing two of the cos(x)'s ( so cos^2(x)) with 1- sin^2(x), multiplying that out then you can integrate to give it in terms of sin^n(x) 's. If that makes sense.
- 12-05-2015 22:04
Expand it to 0.5(4sin^2(x)cos^2(x))cos(x) = 2cos^3(x)sin^2(x)
Then replace cos^2(x) with 1-sin^2(x) = 2cos(x)(1-sin^2(x))sin^2(x)
expand = 2cos(x)(sin^2(x)-sin^4(x)) -
= 2sin^2(x)cos(x) - 2sin^4(x)cos(x)
Then you should be able to use integration with substitution with u = sin(x), so dx = 1/cos(x) du
Giving you the integral of 2u^2 - 2u^4
= 2/3u^3 - 2/5u^5 + c = 2/3sin^3(x) - 2/5sin^5(x) + c
Hopefully that's right haha
- 13-05-2015 15:16