Tính tích phân sau: ∫x sinx/4−cos^2 xdx

Question

Tính tích phân sau:

\(\int _{0}^{\pi }\frac{x\sin x}{4-\cos ^{2} x} dx \)

Answer 1

\(K=\int _{0}^{\pi }\frac{x\sin x}{4-\cos ^{2} x} dx  \)                 

 Đặt \(x=\pi -t\Rightarrow dx=-dt\), ta có:
\(K=-\int _{\pi }^{0}\frac{(\pi -t)\sin (\pi -t)}{4-\cos ^{2} (\pi -t)} dt=\int _{0}^{\pi }\frac{(\pi -t)\sin t}{4-\cos ^{2} t} dt\)   

\(=\int _{0}^{\pi }\frac{\pi \sin t}{4-\cos ^{2} t} dt-K=H-K.\)
\(\Rightarrow 2K=H. \)

\(H=\int _{0}^{\pi }\frac{\pi \sin t}{4-\cos ^{2} t} dt=\pi \int _{0}^{\pi }\frac{d(\cos t)}{(\cos t-2)(\cos t+2)}\)

\( =  \frac{\pi }{4} \int _{0}^{\pi }\left(\frac{1}{\cos t-2} -\frac{1}{\cos t+2} \right)d(\cos t) \)
\(=\frac{\pi }{4} \left. \ln \left|\frac{\cos t-2}{\cos t+2} \right|\right|_{0}^{\pi } =\frac{\pi }{2} \ln 3\). 
Vậy \(K=\frac{\pi }{4} \ln 3.\)