How to Solve the Integration of Sine Problem with Trigonometric Identities

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I am having problem with integration of this
∫sin^3πt

This is what i tried
∫(1-cos^2πt)sinπt
∫sinπt - sinπt(cos^2πt)

∫sinπt - ∫sinπt(cos^2πt)

... and got stuck
OR
∫(1-cos^2πt)sinπt
cos^2t=(1/2)(1+cos2t) so cos^2πt=(1/2)(1+cos2π)
∫((1-(1/2)(1+cos2π))sinπt
∫1/2(sinπt) - (1/2)(cos2π)(sinπt)
and still got stuck
I am not even sure this is the right method to solve that.
I know it should be (cos^3πt)/(3π) - (cosπt)/π but cannot get there

Any help is welcome
 
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greg997 said:
∫sinπt(cos^2πt)
\cos x=t
 
greg997 said:
I am having problem with integration of this
∫sin^3πt dt

This is what i tried
∫(1-cos^2πt)sinπt
∫sinπt - sinπt(cos^2πt)

∫sinπt - ∫sinπt(cos^2πt)

... and got stuck
OR
∫(1-cos^2πt)sinπt
cos^2t=(1/2)(1+cos2t) so cos^2πt=(1/2)(1+cos2π)
∫((1-(1/2)(1+cos2π))sinπt
∫1/2(sinπt) - (1/2)(cos2π)(sinπt)
and still got stuck
I am not even sure this is the right method to solve that.
I know it should be (cos^3πt)/(3π) - (cosπt)/π but cannot get there

Any help is welcome
In my opinion, it's absolutely necessary to include the differential, in this case dt, along with integral symbol.

Which integral are you having difficulty with?
\displaystyle \int\sin(\pi t)\,dt​
or
\displaystyle \int\sin(\pi t)\,\cos^2(\pi t)\,dt\ ?​

For the second one, let u = cos(πt) , then du = _?_
 
Great. That was quite easy. Thank you very much
 
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