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

1 Answer
Jul 9, 2017

#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