Tính tích phân\(\int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\sqrt {1 + \sin x} dx} \).

\[\int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\sqrt {1 + \sin x} dx} = \int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\sqrt {1 + \cos \left( {\frac{\pi }{2} - x} \right)} dx} \]

\[ = \sqrt 2 \int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\sqrt {{{\cos }^2}\left( {\frac{\pi }{4} - \frac{x}{2}} \right)} dx} \]

\[ = \sqrt 2 \int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\left| {\cos \left( {\frac{\pi }{4} - \frac{x}{2}} \right)} \right|dx} \]

Đặt \[u = \frac{\pi }{4} - \frac{x}{2} \Rightarrow du = \frac{{ - 1}}{2}dx\]

Ta được:

I = \(\sqrt 2 \int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\left| {\cos \left( {\frac{\pi }{4} - \frac{x}{2}} \right)} \right|dx} = - = 2\sqrt 2 \int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\left| {\cos u} \right|du} \)

\( = 2\sqrt 2 \int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\left| {\cos u} \right|du} \)

\( = 2\sqrt 2 \int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\cos udu} \)

\( = 2\sqrt 2 \sin u\left| {_{\frac{{ - \pi }}{2}}^{\scriptstyle\frac{\pi }{2}\atop\scriptstyle}} \right.\)

\( = 4\sqrt 2 \).