Evaluate: `int(e^(logx)+sinx)cosxdx`

Answer

`"x sin x" + "cos x" -("cos"2x)/(4)+c`

Explanation

Let `I=int (e^(log x)+sin x)cos x dx`
`=int (x+sin x) cos x dx`
`=int x cos x dx +(1)/(2)int(sin 2x)dx`
`=(x*sin x -int 1*sin x dx)-(cos 2 x)/(4)+c`
`=x sin x + cos x -(cos 2 x)/(4)+c`