Tính tích phân tích phân -pi/2 pi/2 căn bậc hai (1 + sinx) dx
Câu hỏi:
Tính tích phân\(\int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\sqrt {1 + \sin x} dx} \).
Trả lời:
\[\int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\sqrt {1 + \sin x} dx} = \int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\sqrt {1 + \cos \left( {\frac{\pi }{2} - x} \right)} dx} \]
\[ = \sqrt 2 \int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\sqrt {{{\cos }^2}\left( {\frac{\pi }{4} - \frac{x}{2}} \right)} dx} \]
\[ = \sqrt 2 \int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\left| {\cos \left( {\frac{\pi }{4} - \frac{x}{2}} \right)} \right|dx} \]
Đặt \[u = \frac{\pi }{4} - \frac{x}{2} \Rightarrow du = \frac{{ - 1}}{2}dx\]
Ta được:
I = \(\sqrt 2 \int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\left| {\cos \left( {\frac{\pi }{4} - \frac{x}{2}} \right)} \right|dx} = - = 2\sqrt 2 \int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\left| {\cos u} \right|du} \)
\( = 2\sqrt 2 \int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\left| {\cos u} \right|du} \)
\( = 2\sqrt 2 \int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\cos udu} \)
\( = 2\sqrt 2 \sin u\left| {_{\frac{{ - \pi }}{2}}^{\scriptstyle\frac{\pi }{2}\atop\scriptstyle}} \right.\)
\( = 4\sqrt 2 \).