Question

`int_0^(pi/2) dx/(1+2sinx+cosx)`?

Answer 1

`int_0^(pi/2) dx/(1+2sinx+cosx)`

After using `u=tan(x/2)` and `dx=2/(u^2+1)*du` transform, `sinx=(2u)/(u^2+1)` and `cosx=(1-u^2)/(u^2+1)`

Also, boundaries are changed: For `x=0, u=0` and for `x=(pi/2), u=1`

Hence integrand without du became,

`[2/(u^2+1)]/[1+2*(2u)/(u^2+1)+(1-u^2)/(u^2+1)]`

= `[2/(u^2+1)]/[(4u+2)/(u^2+1)]`

= `1/(2u+1)`

Thus, solution of this integral,

`int_0^(pi/2) dx/(1+2sinx+cosx)`

=`int_0^1 1/(2u+1)*du`

=`1/2*Ln(3)-1/2*Ln(1)`

=`1/2*Ln(3)-1/2*0`

=`1/2*Ln(3)`

Explanation:

1) I used u=tan(x/2) transformation

2) I rewrote sinx and cosx in terms of u

3) I changed integral boundaries in terms of u

4) I simplified integrand

5) I took integral and plugged boundaries