Chủ đề này có 1 trả lời

[chiyeuminhem]

$I = \int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{sinx-cosx+1}{sinx+2cosx+3} dx$

[Thai Minh Nhut]

$I = \int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{sinx-cosx+1}{sinx+2cosx+3} dx$

$I = \int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin x-\cos x+1}{\sin x+2\cos x+3} dx$

$\sin x-\cos x +1=A(\sin x+2\cos x+3)+B(-2\sin x+\cos x)+C$

 

$\left\{\begin{matrix} A-2B=1 & & \\ 2A+B=-1 & & \\ 3A+C=1& & \end{matrix}\right. \Rightarrow \left\{\begin{matrix} A=\frac{-1}{5} & & \\ B=\frac{-3}{5} & & \\ C=\frac{8}{5}& & \end{matrix}\right.$

 

$I=\frac{-1}{5}\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\sin x+2\cos x+3}{\sin x+2\cos x+3}dx-\frac{3}{5}\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{-2\sin x+\cos x}{\sin x+2\cos x+3}dx+\frac{8}{5}\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{dx}{\sin x+2\cos x+3} =\frac{-1}{5}\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}dx-\frac{3}{5}\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{d(\sin x+2\cos x+3)}{\sin x+2\cos x+3}+\frac{8}{5}I'$

 

$=\frac{-\pi}{5}-\frac{3}{5}\ln |\sin x+2\cos x+3|_{\frac{-\pi}{2}}^{\frac{\pi}{2}}+\frac{8}{5}I'$

$=\frac{-\pi}{5}-\frac{3}{5}\ln 2+\frac{8}{5}I'$

$I'=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{dx}{\sin x+2\cos x+3}$

 

Đặt $t=\tan {\frac{x}{2}}       \Rightarrow dx=\frac{2dt}{t^2+1}$

 

$\sin x=\frac{2t}{t^2+1}; \cos x=\frac{1-t^2}{1+t^2}$

 

$I'=\int_{-1}^{1}\frac{2dt}{t^2+1}.\frac{1}{\frac{2t}{t^2+1}+2\frac{1-t^2}{1+t^2}+3} $

$=\int_{-1}^{1}\frac{2dt}{2t+2-2t^2+3(t^2+1)} =\int_{-1}^{1}\frac{2dt}{t^2+2t+5} $

$=\int_{-1}^{1}\frac{2dt}{(t+1)^2+2^2}=2.\frac{1}{2}\arctan \frac{t+1}{2}|_{-1}^{1} =\frac{\pi}{4}$

 

$I=\frac{-\pi}{5}-\frac{3}{5}\ln 2+\frac{2\pi}{5}=\frac{\pi}{5}-\frac{3}{5}\ln 2$

Bài viết đã được chỉnh sửa nội dung bởi Thai Minh Nhut: 10-01-2014 - 13:48