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1. Been asked to integrate, and I can't do it.
2. (Original post by Bhaal85)
Been asked to integrate, and I can't do it.
cos2X = 1 - 2sin^2x

^^^Rearrange and integrate^^^
3. (Original post by Leekey)
cos2X = 1 - 2sin^2x

^^^Rearrange and integrate^^^
I knew it was something like that, so its the integral of:

\$ (cos2x-1)/2 dx
4. (Original post by Bhaal85)
I knew it was something like that, so its the integral of:

\$ (cos2x-1)/2 dx
Giving:

(Sin2x)/4 - x/2 + c
5. (Original post by Bhaal85)
I knew it was something like that, so its the integral of:

\$ (cos2x-1)/2 dx
Actually it's (1-cos2x)/2
6. (Original post by Leekey)
cos2X = 1 - 2sin^2x

^^^Rearrange and integrate^^^
or you could say straight away using the half angle identities: sin²x = (1/2) (1 - cos2x)
7. (Original post by mockel)
or you could say straight away using the half angle identities: sin²x = (1/2) (1 - cos2x)
I like to write whole damn thing from start to finish as it proves to the examiner that I actually know something even though my answers are total ****!!!
8. (Original post by shift3)
Actually it's (1-cos2x)/2
Well its official, I've lost it completely.

9. ∫sin²x dx

since cos2x = 1 - 2 sin²x

sin²x = (1/2)(1-cos2x)

so:

∫sin²x dx = (1/2)∫1 - cos2x dx

= x/2 - (1/2).sin2x + C

Hope this helps
10. Well its official, I've lost it completely.
11. (Original post by SUKBarracuda)
I got the right answer, phew.....but initially didn't know how to go about doing it, I can do parts and substitution, its just that it was something I never encountered, now I know. Cheers.
12. (Original post by SUKBarracuda)
∫sin²x dx

since cos2x = 1 - 2 sin²x

sin²x = (1/2)(1-cos2x)

so:

∫sin²x dx = (1/2)∫1 - cos2x dx

= x/2 - (1/2).sin2x + C

Hope this helps
Small error that i thought i'd just point out.. it's x/2 - (1/4).sin2x + C
13. yeah you're right. Just a typo though i'm sure
14. (Original post by SUKBarracuda)
yeah you're right. Just a typo though i'm sure
Yes, damn keyboard.

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Updated: June 2, 2004
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