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int_0^picos^2xsinxdx |
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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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