# Hi,please help me with this integral: integral from 0 to pi [ln(1+sin^2 x)]*cos x * dx. Function is defined in the set of real numbers.

### 1 Answer | Add Yours

First, a different problem:

` int "ln"(1+x^2) dx`

This can be done using integration by parts

`u="ln"(1+x^2)` `dv=dx`

`du=(1)/(1+x^2)2x dx` `v=x`

`int "ln"(1+x^2) dx = x"ln"(1+x^2)-int (2x^2)/(1+x^2) dx `

`=x"ln"(1+x^2)-2 int (x^2+1-1)/(1+x^2) dx`

`=x"ln"(1+x^2)-2 int 1 -(1)/(1+x^2) dx`

`=x"ln"(1+x^2)-2x + 2 int (1)/(1+x^2) dx`

`=x"ln"(1+x^2)-2x + 2 "Tan"^(-1) x +C`

Back to the original problem:

Try a substitution:

`u="sin"x`

`du="cos"x dx`

But `"sin" x` is not a 1-1 function on `[0,pi]`

`"sin"x` is symmetric about `(pi)/2`

So `int_0^pi = 2 int_0^((pi)/(2))`

making the substitution changes the bounds:

`x=0 => u=0`

`x=(pi)/(2) => u=1`

`2 int_0^1 ln(1+u^2) du`

Using the first calculation:

`2 int_0^1 ln(1+u^2)du`

`=2 [u"ln"(1+u^2)-2u + 2 "Tan"^(-1) u |_0^1 ]`

`=2[ "ln"(2)-2+2"Tan"^(-1) (1) - 0 + 0 -2 "Tan"^(-1)0 ]`

`=2[ "ln"(2) -2 +2(pi)/(4) ]`

`=2 "ln" 2 - 4 + pi`