Question

How to integrate `int1/(sinx+tanx)dx`?

Answer 1

`int \ 1/(sinx + tanx) \ dx = 1/4 ln|(cosx-1)/(cosx+1)| -1/2 1/(cosx+1) +C`

Explanation:

Let:

`I = int \ 1/(sinx + tanx) \ dx`

Which we can write as follows:

`I = int \ 1/(sinx + sinx/cosx) \ dx`
`\ \ = int \ 1/((sinxcosx + sinx)/cosx) \ dx`
`\ \ = int \ cosx/(sinx(cosx + 1)) \ dx`
`\ \ = int \ (sinxcosx)/(sin^2x(1+cosx)) \ dx`
`\ \ = int \ (sinxcosx)/((1-cos^2x)(1+cosx)) \ dx`
`\ \ = int \ (sinxcosx)/((1-cosx)(1+cosx)(1+cosx)) \ dx`
`\ \ = int \ (sinxcosx)/((1-cosx)(1+cosx)^2) \ dx`

Now we can perform a trig substitution; Let

`u=cosx => (du)/dx=-sinx`

Substituting into our integral we get:

`I = int \ ((-1)(u))/((1-u)(u+1)^2) \ du`
`\ \ = int \ (u)/((u-1)(u+1)^2) \ du`

Which we can decompose into partial fraction; as:

`(u)/((u-1)(u+1)^2) -= A/(u-1) + B/(u+1)+C/(u+1)^2`
`" " = (A(u+1)^2 + B(u-1)(u+1)+C(u-1))/((u-1)(u+1)^2)`

Leading to:

`u = A(u+1)^2 + B(u-1)(u+1)+C(u-1)`

Put `u=1 => 1=4A => A=1/4`
Put `u=-1 => -1=-2C=>C=1/2`

Comparing Coefficients:

`"Coeff"(u^2) => 0=A+B=>B=-1/4`

Hence the partial fraction decomposition gives us:

`I = int \ (1/4)/(u-1) - (1/4)/(u+1)+(1/2)/(u+1)^2 \ du`
`\ \ = 1/4 \ int \ 1/(u-1) \ du - 1/4 \ int 1/(u+1) \ du +1/2 \ int 1/(u+1)^2 \ du`
`\ \ = 1/4 ln|u-1| - 1/4 ln|u+1| +1/2 (-1)/(u+1) +C`
`\ \ = 1/4 ln|(u-1)/(u+1)| -1/2 1/(u+1) +C`

Restoring the substitution we get:

`I = 1/4 ln|(cosx-1)/(cosx+1)| -1/2 1/(cosx+1) +C`