1 reply to this topic

[Sonhai224]

Tính

$\int_{-\pi/2}^{\frac{\pi}{2}}\frac{x+cosx}{4-sin^2x}dx$

Edited by Sonhai224, 03-01-2018 - 15:51.

[conanthamtulungdanhkudo]

Tính

$\int_{-\pi/2}^{\frac{\pi}{2}}\frac{x+cosx}{4-sin^2x}dx$

$\int_{\frac{-\pi }{2}}^{\frac{\pi }{2}}\frac{x}{4-sin^2x}$$+\int_{\frac{-\pi }{2}}^{\frac{\pi }{2}}\frac{cosxdx}{4-sin^2x}$

Đặt tích phân đầu $=I$

Tìm $I$

Đặt $x=-t$$\Rightarrow dx=-dt$

$x=\frac{\pi }{2}\Rightarrow t=\frac{-\pi }{2}$

$x=\frac{-\pi }{2}\Rightarrow t=\frac{\pi }{2}$

$I=\int_{\frac{\pi }{2}}^{\frac{-\pi }{2}}\frac{tdt}{4-sin^2t}=-I$

$\Rightarrow 2I=0\Rightarrow I=0$

Phần sau $\int_{\frac{-\pi }{2}}^{\frac{\pi }{2}}\frac{d(sinx)}{4-sin^2x}=\frac{1}{4}ln\left | \frac{2+sinx}{sinx-2} \right |$(thế cận từ $\frac{-\pi }{2}\rightarrow \frac{\pi }{2}$