$\int_{0}^{\frac{π}{2}} \frac{d x}{1 + 2 sin x + cos x}$ $int_0^(pi/2) dx/(1+2sinx+cosx)$ ?

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$\int_{0}^{\frac{π}{2}} \frac{d x}{1 + 2 sin x + cos x}$ $int_0^(pi/2) dx/(1+2sinx+cosx)$ ?

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Answer 1 · 2017-07-09T02:12:20

$\int_{0}^{\frac{π}{2}} \frac{d x}{1 + 2 sin x + cos x}$ $int_0^(pi/2) dx/(1+2sinx+cosx)$

After using $u = tan (\frac{x}{2})$ $u=tan(x/2)$ and $d x = \frac{2}{u^{2} + 1} \cdot d u$ $dx=2/(u^2+1)*du$ transform, $sin x = \frac{2 u}{u^{2} + 1}$ $sinx=(2u)/(u^2+1)$ and $cos x = \frac{1 - u^{2}}{u^{2} + 1}$ $cosx=(1-u^2)/(u^2+1)$

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

Hence integrand without du became,

$\frac{\frac{2}{u^{2} + 1}}{1 + 2 \cdot \frac{2 u}{u^{2} + 1} + \frac{1 - u^{2}}{u^{2} + 1}}$ $[2/(u^2+1)]/[1+2*(2u)/(u^2+1)+(1-u^2)/(u^2+1)]$

= $\frac{\frac{2}{u^{2} + 1}}{\frac{4 u + 2}{u^{2} + 1}}$ $[2/(u^2+1)]/[(4u+2)/(u^2+1)]$

= $\frac{1}{2 u + 1}$ $1/(2u+1)$

Thus, solution of this integral,

$\int_{0}^{\frac{π}{2}} \frac{d x}{1 + 2 sin x + cos x}$ $int_0^(pi/2) dx/(1+2sinx+cosx)$

= $\int_{0}^{1} \frac{1}{2 u + 1} \cdot d u$ $int_0^1 1/(2u+1)*du$

= $\frac{1}{2} \cdot ln (3) - \frac{1}{2} \cdot ln (1)$ $1/2*Ln(3)-1/2*Ln(1)$

= $\frac{1}{2} \cdot ln (3) - \frac{1}{2} \cdot 0$ $1/2*Ln(3)-1/2*0$

= $\frac{1}{2} \cdot ln (3)$ $1/2*Ln(3)$

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

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$\int_{0}^{\frac{π}{2}} \frac{d x}{1 + 2 sin x + cos x}$ $int_0^(pi/2) dx/(1+2sinx+cosx)$ ?

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