Question

How do you Integrate `cosx/(sinx)^2+sinx`?

Answer 1

The required integral is

`I = int[cosx/(sinx)^2 + sinx]dx`

This can be evaluated as two separate integrals,

`I = I_1 + I_2` , where `I_1 = intcosx/(sinx)^2dx` and `I_2 = intsinxdx`

The solution of `I_2` is trivial,

`I_2 = intsinxdx = -cosx + C_2`

For `I_1:`

Let `sinx = t`

Differentiating with respect to t,

`cosxdx/dt = 1`

`=> cosxdx = dt`

This transforms `I_1` to

`I_1 = int1/t^2dt`

which has the simple solution of

`I_1 = -1/t + C_1`

Replacing the value of `t = sinx`

`I_1 = -1/sinx + C_1`

Combining both solutions,

`I = I_1 + I_2`
`=> I = -1/sinx + C_1 -cosx + C_2`

The constants of integration `C_1` and `C_2` can be added without affecting the solution.

Finally, the answer is

`I = -1/sinx -cosx + C`