You are Here: Home >< Maths

# ∫ sin^2 x cos^3 x dx?? watch

1. I don't know what to do!

Anyone done one like this before?
cheers
2. Ok

sin^2x Cos^3x = sin^2x.CosX.(Cos^2X)
= sin^2x.CosX.(1-sin^2X)
= sin^2x. Cosx - sin^4x.cosx

Then you can integrate that as the product pairs have one as the differential of the other
3. (Original post by Mysticmin)
Ok

sin^2x Cos^3x = sin^2x.CosX.(Cos^2X)
= sin^2x.CosX.(1-sin^2X)
= sin^2x. Cosx - sin^4x.cosx

Then you can integrate that as the product pairs have one as the differential of the other

ahh thankie!
how did you know to do that?
4. (Original post by kimoni)
ahh thankie!
how did you know to do that?
you have to split the cos^3x up into cos^2X * cosx. Then you use the identity cos^2x + sin^2x = 1, rearrange and you get cos^2x = 1 - sin^2x. Then subsitutute that in for cos^2x in the integration equation.

5. (Original post by Mysticmin)
tomorrow!
6. (Original post by kimoni)
tomorrow!
lol, good luck, (p1?)

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: June 8, 2004
Today on TSR

### University open days

1. University of Cambridge
Wed, 26 Sep '18
2. Norwich University of the Arts
Fri, 28 Sep '18
3. Edge Hill University
Faculty of Health and Social Care Undergraduate
Sat, 29 Sep '18
Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams