1.

Evaluate : (i) `intxsec^(2)xdx` (ii) `intxsin2xdx`

Answer» (i) Integrating by parts, taking x as the first function, we have
`intxsec^(2)xdx=x*intsec^(2)xdx-int{(d)/(dx)(x)*intsec^(2)xdx}dx`
`=xtanx-int1*tanxdxtanx+log|cosx|+C`.
(ii) Integrating by parts, taking x as the first function, we get
`intxsin2x=x*intsin2xdx-int{(d)/(dx)(x)*intsin2xdx}dx`
`x*((-cos2x)/(2))-int1*((-cos2x)/(2))dx`
`=(-xcos2x)/(2)+(1)/(2)intcos2xdx`
`=(-xcos2x)/(2)+(1)/(2)*(sin2x)/(2)+C`
`(-xcos2x)/(2)+(1)/(4)sin2x+C`.


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