Original post by Ganhad
ingtegral of sin^2x dx how would i do this ?
Consider an identity for cos(2x)
Original post by Abed1993
Do you know how to integrate sin x?
This will pay no part in the question.
Original post by Ganhad
ingtegral of sin^2x dx how would i do this ?
OP recall that 1-2sin^2x = cos(2x)
(edited 12 years ago)
Sin ^2 x ... becomes 0.5(1-cos2x)
Then intergrate that ...
Then intergrate that ...
Original post by Ganhad
cos2a=2cos^2a-1 ?
yes but perhaps consider the RHS of the identity with sine functions instead of cosine functions
Original post by Extricated
This will pay no part in the question.
OP recall that 1+2sin^2x = cos(2x)
OP recall that 1+2sin^2x = cos(2x)
1-2sin^2x
Original post by Ganhad
how does that relate to this im confused if you can explain the link i would get it much better like the first 2 steps i have to do 
?
?You don't know how to integrate the original function but you can turn it into functions which you can integrate, no?
Original post by Ganhad
how does that relate to this im confused if you can explain the link i would get it much better like the first 2 steps i have to do ?
?e.g.
Original post by Ganhad
how does that relate to this im confused if you can explain the link i would get it much better like the first 2 steps i have to do ?
?Using trig rules we know that 1-2sin^2x = cos(2x)
Rearrange this to get sin^2x on the LHS so:
1 = cos(2x) + 2sin^2x
1-cos(2x) = 2sin^2x
sin^2x = [1-cos(2x)]/2
Therefore ∫ sin^2x = ∫ [1-cos(2x)]/2
We can integrate ∫ [1-cos(2x)-1]/2
- We can drag the 1/2 out to give 0.5∫1-cos(2x)
- Integrating then gives 0.5[x-0.5sin(2x)]
- Multiply out brackets gives 0.5x-0.25sin(2x)
∫ sin^2x = 0.5x - 0.25sin(2x) + c
(edited 12 years ago)
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