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

Jul 9, 2017

${\int}_{0}^{\frac{\pi}{2}} \frac{\mathrm{dx}}{1 + 2 \sin x + \cos x}$

After using $u = \tan \left(\frac{x}{2}\right)$ and $\mathrm{dx} = \frac{2}{{u}^{2} + 1} \cdot \mathrm{du}$ transform, $\sin x = \frac{2 u}{{u}^{2} + 1}$ and $\cos x = \frac{1 - {u}^{2}}{{u}^{2} + 1}$

Also, boundaries are changed: For $x = 0 , u = 0$ and for $x = \left(\frac{\pi}{2}\right) , u = 1$

Hence integrand without du became,

$\frac{\frac{2}{{u}^{2} + 1}}{1 + 2 \cdot \frac{2 u}{{u}^{2} + 1} + \frac{1 - {u}^{2}}{{u}^{2} + 1}}$

= $\frac{\frac{2}{{u}^{2} + 1}}{\frac{4 u + 2}{{u}^{2} + 1}}$

= $\frac{1}{2 u + 1}$

Thus, solution of this integral,

${\int}_{0}^{\frac{\pi}{2}} \frac{\mathrm{dx}}{1 + 2 \sin x + \cos x}$

=${\int}_{0}^{1} \frac{1}{2 u + 1} \cdot \mathrm{du}$

=$\frac{1}{2} \cdot L n \left(3\right) - \frac{1}{2} \cdot L n \left(1\right)$

=$\frac{1}{2} \cdot L n \left(3\right) - \frac{1}{2} \cdot 0$

=$\frac{1}{2} \cdot L n \left(3\right)$

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