Question

What is `int tan(2x-3)-cot(3x)`?

Answer 1

-1/2ln(cos(2x-3)) - 1/3ln(sin(3x)) + C

Explanation:

`I= int(tan(2x-3) - cot(3x))dx`
`I =inttan(2x-3)dx - int(cot(3x)dx`
`I =int sin(2x-3)/cos(2x-3)dx - int cos(3x)/sin(3x)dx`
`I=I_1 - I_2`
Let us name the two integrals as `I_1` and `I_2`
and solve them separately.

`I_1 = int sin(2x-3)/cos(2x-3) dx`

Let us solve this first

Let `u=cos(2x-3)`
Diffrentiating with respect to `x` using chain rule.

`(du)/dx = -sin(2x-3)d/dx(2x-3)`
`(du)/dx = -sin(2x-3)(2)`
`(du)/dx = -2sin(2x-3)`
`du = -2sin(2x-3)dx`
`(du)/-2 = sin(2x-3)dx`
`I_1 = int 1/u ((du)/-2)`
`I_1 = -1/2int (du)/u`
`I_1 = -1/2ln(u)`
`I_1 = -1/2 ln(cos(2x-3))`

`I_2 = intcos(3x)/sin(3x) dx`
Let `u=sin(3x)`
Differentiating with respect to `x` using chain rule.

`(du)/dx = cos(3x)d/dx(3x)`
`(du)/dx = cos(3x)3`
`(du)/dx =3cos(3x)`
`(du)/3 = cos(3x)dx`

`I_2 = int1/u ((du)/3)`
`I_2 = 1/3 int (du)/u`
`I_2 = 1/3 ln(u)`

`I_2 = 1/3 ln(sin(3x))`

We had got till the step `I=I_1-I_2`

`-1/2ln(cos(2x-3)) - 1/3ln(sin(3x)) + C` Answer