Question

Find `int \ (sin2x+1)/(1-sin2x) \ dx`?

Answer 1

`int \ (sin2x+1)/(1-sin2x) \ dx = - x -2/(tanx-1) + C`

Explanation:

We seek:

`I = int \ (sin2x+1)/(1-sin2x) \ dx`

We can write this as:

`I = int \ (1+sin2x)/(1-sin2x) \ dx`

`\ \ \ = -int \ (-1-sin2x)/(1-sin2x) \ dx`

`\ \ \ = -int \ (1-sin2x-2)/(1-sin2x) \ dx`

`\ \ \ = -int \ 1 -2/(1-sin2x) \ dx`

`\ \ \ = - int \ dx + 2 \ int 1/(1-sin2x) \ dx` ..... [A]

The first integral is standard, and for the second we have to perform some further analysis:

Using a standard Weierstraß trigonometric identity we have:

`\ \ \ \ \ \ \ sin x = (2tan(x/2))/(1+tan^2(x/2))`

`:. sin 2x = (2tan(x))/(1+tan^2(x)) = 2tanx/sec^2x`

So, for the second integral, `I_2`, say,we have using this identity:

`I_2 = int \ 1/(1-sin2x) \ dx`

`\ \ \ = int \ 1/(1-2tanx/sec^2x) \ dx`

`\ \ \ = int \ 1/( (sec^2x-2tanx)/sec^2x) \ dx`

`\ \ \ = int \ (sec^2x)/( sec^2x-2tanx) \ dx`

`\ \ \ = int \ (sec^2x)/( tan^2x+1-2tanx) \ dx`

`\ \ \ = int \ (sec^2x)/( tanx-1 )^2 \ dx`

We can now easily integrate this if we perform a substitution:

Let `u= tanx-1 => (du)/dx = sec^2x`

Substituting into `I_2` we have:

`I_2 = int \ 1/u^2 \ du`
`\ \ \ = -1/u`
`\ \ \ = -1/(tanx-1)`

Hence, we can now integrate [A] to get:

`I = - x + 2(-1/(tanx-1)) + C`
`\ \ = - x -2/(tanx-1) + C`