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Integrating sin squared x

why can't you make sinx=u and use this substitution to do it?
Original post by runny4
why can't you make sinx=u and use this substitution to do it?


Well, you'd get

du=cos⁡x dxdu = \cos x \ dx

There's the problem.
Hint: You may wish to consider the following trigonometric identity.

cos⁡(2x)=1−2sin⁡2(x)\displaystyle \cos (2x)=1-2\sin ^2(x).
Original post by Thediplomat56
image.jpg

Sorry it's meant to be -1/4sin2theta


Don't post full solutions.
Original post by Thediplomat56
image.jpg

Sorry it's meant to be -1/4sin2theta


Edit: WHoops thought you were the OP. Stupid 16Characters.
Reply 5
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davros
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Original post by runny4
why can't you make sinx=u and use this substitution to do it?


Well, what do you end up with when you try this? :smile:
Reply 6
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runny4
OP
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Original post by Indeterminate
Well, you'd get

du=cos⁡x dxdu = \cos x \ dx

There's the problem.


why is that a problem?- you end up with something in terms of sin and cos
Reply 7
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runny4
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Original post by davros
Well, what do you end up with when you try this? :smile:


you end up with sin cubed x over 3cosx
Original post by runny4
you end up with sin cubed x over 3cosx


Presumably you had an integral of

∫u2cosxdu \displaystyle \int \frac{u^2}{cos x} du

The issue is that cosx=cos(arcsin(u)) cos x = cos(arcsin (u)) so you cannot treat it as a constant when integrating w.r.t u.
(edited 8 years ago)
Reply 9
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runny4
OP
6
Original post by 16Characters....
Presumably you had an integral of

∫u2cosxdu \displaystyle \int \frac{u^2}{cos x} du

The issue is that cosx=cos(arcsin(u)) cos x = cos(arcsin (u)) so you cannot treat it as a constant when integrating w.r.t u.


ok thank you i thought that you could treat cosx as constant with respect to u and bring it to the front of the integral but obviously u can't.

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