Example Evaluate the following the definite integral．

$$\int_{0}^{\frac{\pi}{2}}\frac{\sin{x}}{\sqrt{1+\cos{x}}} dx$$

Answer We note that every trig integral can be expressed by rational function.

1.	The integral of a sum is the sum of the integrals． $\int (f(x) \pm g(x))dx = \int f(x)dx \pm \int g(x)dx$
2.	The integral of a constant multiple is a constant times the integrals． $\int_{a}^{b} (\alpha f(x))dx = \alpha \int_{a}^{b} f(x)$
3.	Integration By Substitution． $\int_{a}^{b} f(x)dx = \int_{\Phi^{-1}(a)}^{\Phi^{-1}(b)} f(\Phi(t))(\Phi(t))'dt$
4.	Integration By Parts. $\int_{a}^{b} f(x)g'(x)dx = f(x)g(x)]_{a}^{b} - \int_{a}^{b} f'(x)g(x)dx$
5.	Partial Fraction Expansion． $\frac{3x+2}{x(x^2+1)} = \frac{A}{x} + \frac{Bx+C}{x^2+1}$
6.	Basic Integration Formula． $\int_{a}^{b} e^{x}dx = e^{x}]_{a}^{b}, \int_{a}^{b} \sin{x}dx = -\cos{x}]_{a}^{b}, \int_{a}^{b} \frac{1}{x}dx = \log\vert x\vert]_{a}^{b}$

Put $t = \sqrt{1 + \cos{x}}$. Then $t^2 = 1 + \cos{x}$ implies $2tdt = -\sin{x}dx$．Then the interval of integration chages.

$$\begin{array}{llll} x&0 &\Rightarrow & \frac{\pi}{2}\\ \hline t&\sqrt{2} &\Rightarrow & 1 \end{array}$$

Thus,

$$\int_{0}^{\frac{\pi}{2}}{\frac{\sin{x}}{\sqrt{1 + \cos{x}}}}\ dx = \int_{\sqrt{2}}^{1}{\frac{-2tdt}{t}}$$

$$= -2\int_{\sqrt{2}}^{1}\ dt = -2t\mid_{\sqrt{2}}^{1} = -2(1 - \sqrt{2}) = 2(\sqrt{2} - 1)$$