Evaluate :
`(i) int_(0)^(pi//2)xcosxdx` `(ii) int_(0)^(pi)cos2xlog sinx dx`
`(iii) int_(1)^(2)(logx)/(x^(2))dx` `(iv) int_(0)^(pi//6)(2+3x^(2))cos3x dx`

Explanation

`(i)` Integrating by parts, we get
` int_(0)^(pi//2)xcosxdx=[xsinx]_(0)^(pi//2)-int_(0)^(pi//2)1*sinxdx`
`=(pi)/(2)+[cosx]_(0)^(pi//2)=((pi)/(2)-1)`.
`(ii)` Integrating by parts, taking `log(sinx)` as the first function, we get
` int_(0)^(pi)cos2xlog sinx dx`
`=[(logsinx)*(sin2x)/(2)]_(0)^(pi)-int_(0)^(pi)(cotx*(sin2x)/(2))dx`
`=0-int_(0)^(pi)(cosx)/(sinx)*(2sinxcosx)/(2)dx=-int_(0)^(pi)cos^(2)xdx`
`=-(1)/(2)int_(0)^(pi)2cos^(2)xdx=-(1)/(2)int_(0)^(pi)(1+cos2x)dx`
`=-(1)/(2)*[x+(sin2x)/(2)]_(0)^(pi)=-(pi)/(2)`.
`(iii)` Integrating by parts, taking `(logx)` as the first function, we get
` int_(1)^(2)(logx)/(x^(2))dx=int_(1)^(2)(logx)*x^(-2)dx`
`=[(logx)(-(1)/(x))]_(1)^(2)-int_(1)^(2)(1)/(x)*(-(1)/(x))dx`
`=[-(log2)/(2)+(log1)/(1)]+int_(1)^(2)(dx)/(x^(2))`
`=(-log2)/(2)-[(1)/(x)]_(1)^(2)=(-log2)/(2)-{(1)/(2)-1}=((1-log2)/(2))`.
`(iv) int_(0)^(pi//6)(2+3x^(2))cos3x dx`
`=2int_(0)^(pi//6)cos3xdx+3int_(0)^(pi//6)x^(2)cos3xdx`
`=2[(sin3x)/(3)]_(0)^(pi//6)+3{[x^(2)((sin3x)/(3))]_(0)^(pi//6)-int_(0)^(pi//6)2x((sin3x)/(3))dx}` [integrating by parts]
`=(2)/(3)+(pi^(2))/(36)-2int_(0)^(pi//6)xsin3xdx`
`=(2)/(3)+(pi^(2))/(36)-2{[x((-cos3x)/(3))]_(0)^(pi//6)-int_(0)^(pi//6)1*((-cos3x)/(3))dx}` [integrating by parts]
`=(2)/(3)+(pi^(2))/(36)+(2)/(3)[xcos3x]_(0)^(pi//6)-(2)/(3)*[(sin3x)/(3)]_(0)^(pi//6)`
`=(2)/(3)+(pi^(2))/(36)-(2)/(9)((pi^(2))/(36)+(4)/(9))=(1)/(36)(pi^(2)+16)`.