Bài 105406

Question

Tính tích phân: $\int\limits_0^{\frac{\pi }{2}} {\frac{\cos xdx}{{\sqrt {1 + cos^2x} }}} $

Tiểu Bắc · Answer 1 · 5/23/2012 4:15:52 PM

$\int\limits_0^{\frac{\pi }{2}} {\frac{\cos xdx}{{\sqrt {1 + cos^2x} }}} =\int\limits_0^{\frac{\pi }{2}} {\frac{dsinx}{{\sqrt {2-sin^2x} }}} =\int\limits_0^{\frac{\pi }{2}} {\frac{dsinx}{{\sqrt {2}.\sqrt{1-\frac{sin^2x}{2} } }}} $
$=\int\limits_0^{\frac{\pi }{2}} {\frac{d(\frac{sinx}{\sqrt{2} } )}{{\sqrt {1-(\frac{sinx}{\sqrt{2} } )^2} }}} =\int\limits_{0}^{\frac{1}{\sqrt{2} } } \frac{dt}{\sqrt{1-t^2} } (t=\frac{sinx}{\sqrt{2} } )$

Đặt $t=\sin u$

$\Rightarrow I=\int\limits^{\frac{\pi}{4}}_0\frac{\cos udu}{\cos u}=\int\limits^{\frac{\pi}{4}}_0 du=u|^{\frac{\pi}{4}}_0=\frac{\pi}{4}$

Đăng bài 23-05-12 04:15 PM

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