[Toán 12] Tính nguyên hàm

V

vivietnam

$I=\int \dfrac{sinxdx}{cosx\sqrt{1+sin^2x}}$
Đặt $cosx=t \Longrightarrow sinxdx=-dt$
$I=-\int \dfrac{dt}{t.\sqrt{2-t^2}}=-\dfrac{1}{2}\int \dfrac{d(t^2)}{t^2\sqrt{2-t^2}}$
Đặt $\sqrt{2-t^2}=u \Longrightarrow -d(t^2)=2udu$
$I=\int \dfrac{udu}{(u^2+2).u}=\int \dfrac{du}{2+u^2}=\dfrac{1}{\sqrt{2}}arctan(\dfrac{u}{\sqrt{2}})$
 
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