Integrate the functions`e^(2x)sinx`

Explanation

`" Let " I= int e^(2x) sin x dx`
` rArr I= sin x. int e^(2x) dx`
` - int [(d)/(dx) (sin x). int e^(2x) dx]dx`
`=sin x. (e^(2x))/(2)- int cos x. (e^(2x))/(2) dx`
` =sin x.(e^(2x))/(2)-(1)/(2) int e^(2x) cos x dx`
`rArr I= sin x.(e^(x))/(2)-(1)/(2) I_(1) " "......(1)`
`" where " I_(1)= int e^(2x) cos x dx `
` rArr I_(1) = cos x . int e^(2x) dx`
` - int [(d)/(dx) (cos x). int e^(2x) dx ]dx`
`=cos x.(e^(2x))/(2)-int (-sin x). (e^(2x))/(2) dx+C_(1)`
`=(e^(2x)cos x)/(2) +(1)/(2) int e^(2x) sin x dx +C_(1)`
Put the value of `I_(1)` in equation (1)
`I= (e^(2x) sin x)/(2)`
`-(1)/(2)((e^(2x)cosx)/(2)+(1)/(2) int e^(2x) sin x dx +C_(1))`
`rArr int e^(2x) sin x dx =(e^(2x) sin x)/(2)-(e^(2x)cos x)/(4)-(1)/(2) C_(1)`
` rArr I+(I)/(4) =(e^(2x)sinx)/(2)-(e^(2x)cosx)/(4)-(1)/(2)C_(1)`
`rArr (5)/(4) I= (e^(2x) sin x)/(2) -(e^(2x) cos x)/(4) -(1)/(2) C_(1)`
`rArr I=(4)/(4) ((e^(2x) sinx)/(2)-(e^(2x)cos x)/(4)-(1)/(2) C_(1))`
`rArr int e^(2x) sin x dx`
` =(2)/(5) e^(2x) sin x-(1)/(5) e^(2x) cos x-(2)/(5) C_(1)`
`=(2)/(5) e^(2x) sin x-(1)/(5) e^(2x) cos x+c`
`" where " C=-(2)/(5) C_(1)`