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int_0^picos^2xsinxdx

Answer»

SOLUTION :`int_0^picos^2x.sinxdx`
[PUT cosx=t, Then sindx=-DT
When x=0, t=1, When `x=pi`, t=-1]
=`int_1^-1t^2(-dt)=int_-1^1t^2dt=[t^3/3]_-1^1`
=1/3+1/3=2/3


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