Evaluate following integrals,
`(i) intlogxdx`
`(ii) inte^(x)cosxdx`

Explanation

`(i) l=intlogxdx`
Note : For such questions where we are not able to find a function whose derivative is `logx`, we consider `logx` as first function and unity as second function.
`:.l=funderset(II)((1))underset(I)((logx))dx`
`:.l=logxintdx-int{(d)/(dx)logxintdx}dx`
`:.l=xlogx-int(1)/(x)*xdx`
`impliesl=xlogx-x+C`
`(ii) l=intunderset(I)(e^(x))*underset(II)(cosx)dx`
`l=e^(x)intcosxdx-int{(d)/(dx)e^(x)*intcosxdx}dx`
`:.l=e^(x)sinx-intunderset(I)(e^(x))underset(II)(sinx)dx` [Again using By parts]
`:.l=e^(x)sinx-[e^(x)intsinxdx-int{(d)/(dx)e^(x)*intsinxdx}dx]`
`:.l=e^(x)sinx-[-e^(x)cosx-inte^(x)(-cosx)dx]`
`:.l=e^(x)sinx+e^(x)cosx-inte^(x)coxdx`
`:.2l=e^(x)sinx+e^(x)cosx` (As `inte^(x)cosxdx=l`)
`impliesl=(e^(x))/(2)(sinx+cosx)+C`