Tính tích phân $I=\displaystyle \int \limits_{\frac{\pi}{4}}^{\frac{\pi}{3}} \dfrac{\sin x}{\cos^2x\sqrt{2-\sin^2x}}dx$.

Tính tích phân $I=\displaystyle \int \limits_{\frac{\pi}{4}}^{\frac{\pi}{3}} \dfrac{\sin x}{\cos^2x\sqrt{2-\sin^2x}}dx$.

Giải

$\begin{align} I &=\displaystyle \int \limits_{\frac{\pi}{4}}^{\frac{\pi}{3}} \dfrac{\sin x}{\frac{\cos x}{\cos x}\sqrt{1+\cos^2x}}.\dfrac{1}{\cos^2 x}dx \\ &=\displaystyle \int \limits_{\frac{\pi}{4}}^{\frac{\pi}{3}} \dfrac{\tan x}{\sqrt{\dfrac{1}{\cos ^2x}+1}}.\dfrac{1}{\cos^2 x}dx \\ &= \displaystyle \int \limits_{\frac{\pi}{4}}^{\frac{\pi}{3}} \dfrac{\tan x}{\sqrt{\tan^2x+2}}.\dfrac{1}{\cos^2 x}dx \end{align}$

Đặt $t=\sqrt{\tan^2x+2} \Rightarrow tdt=\dfrac{\tan x}{\cos^2 x}dx$

$x=\dfrac{\pi}{4} \Rightarrow t=\sqrt{3} , \,\, x=\dfrac{\pi}{3} \Rightarrow t=\sqrt{5}$

$ I =\displaystyle \int \limits_{\sqrt{3}}^{\sqrt{5}} dt= \sqrt{5}-\sqrt{3}$.