Question

How do I find the integral `inte^(2x)*sin(3x)dx` ?

Answer 1

`=3/13e^(2x)(2/3sin(3x)-cos(3x))`

Explanation :

`I=inte^(2x)*sin(3x)dx` .........`(i)`

Using Integration by Parts,

`I=e^(2x)intsin(3x)dx-int(d/dx(e^(2x))intsin(3x)dx)dx`

`I=e^(2x)(-cos(3x)/3)-int2*e^(2x)(-cos(3x)/3)dx`

`I=-e^(2x)*cos(3x)/3+2/3inte^(2x)cos(3x)dx`

`I=-e^(2x)*cos(3x)/3+2/3I_1` .........`(ii)`

where, `I_1=inte^(2x)cos(3x)dx`

Again, using Integration by Parts,

`I_1=e^(2x)intcos(3x)-int(d/dx(e^(2x))intcos(3x)dx)dx`

`I_1=e^(2x)*sin(3x)/3-int(2*e^(2x)sin(3x)/3)dx`

`I_1=e^(2x)*sin(3x)/3-2/3inte^(2x)sin(3x)dx`

`I_1=e^(2x)*sin(3x)/3-2/3I`, [using `(i)`]

now, putting `I_1` in `(ii)` yields,

`I=-e^(2x)*cos(3x)/3+2/3(e^(2x)*sin(3x)/3-2/3I)`

`I=-e^(2x)*cos(3x)/3+2/9e^(2x)*sin(3x)-4/9I`

`I+4/9I=-e^(2x)*cos(3x)/3+2/9e^(2x)*sin(3x)`

`13/9I=e^(2x)/3(2/3sin(3x)-cos(3x))`

`I=3/13e^(2x)(2/3sin(3x)-cos(3x))`