Tính tích phân sau: ∫x+cos x/4−sin^xdx

Question

Tính tích phân sau:

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

Answer 1

Ta có: \(J=\int _{-\frac{\pi }{2} }^{\frac{\pi }{2} }\frac{x}{4-\sin ^{2} x} dx+\int _{-\frac{\pi }{2} }^{\frac{\pi }{2} }\frac{\cos x}{4-\sin ^{2} x} dx=J_{1} +J_{2} .\)

Ta có: \(f(x)=\frac{x}{4-\sin ^{2} x}\)  là hàm lẻ trên \(\left[-\frac{\pi }{2} ;\frac{\pi }{2} \right],\)

do đó: \(J_{1} =0.\)

\(J_{2} =-\int _{-\frac{\pi }{2} }^{\frac{\pi }{2} }\frac{d(\sin x)}{(\sin x-2)(\sin x+2)}\) \(=-\frac{1}{4}  \int _{-\frac{\pi }{2} }^{\frac{\pi }{2} }\left(\frac{1}{\sin x-2} -\frac{1}{\sin x+2} \right)d(\sin x)\)

\(=-\frac{1}{4} \left. \ln \left|\frac{\sin x-2}{\sin x+2} \right|\right|_{-\frac{\pi }{2} }^{\frac{\pi }{2} } =\frac{1}{2} \ln 3 \)

\(Vậy J=\frac{1}{2} \ln 3.\)