Tính tích phân sau: ∫sinx/(1+sin2x)dx

Question

Tính tích phân sau:

\(\int _{0}^{\frac{\pi }{2} }\frac{\sin x}{1+\sin 2x} {\rm d}x \)

Answer 1

\(I=\int _{0}^{\frac{\pi }{2} }\frac{\sin x}{1+\sin 2x} {\rm d}x \)

 Đặt \(t=\frac{\pi }{2} -x\Rightarrow {\rm d}t=-{\rm d}x.\)  

Đổi cận: \(x=\frac{\pi }{2} \Rightarrow t=0;\, x=0\Rightarrow t=\frac{\pi }{2} \)

 Khi đó: \(I=-\int _{\frac{\pi }{2} }^{0}\frac{\sin \left(\frac{\pi }{2} -t\right)x}{1+\sin \left(\pi -2t\right)} {\rm d}t=\int _{0}^{\frac{\pi }{2} }\frac{\cos t}{1+\sin 2t} {\rm d}t  \)

 Suy ra: \(2I=\int _{0}^{\frac{\pi }{2} }\frac{\sin x+\cos x}{1+\sin 2x} {\rm d}x=\int _{0}^{\frac{\pi }{2} }\frac{\sin x+\cos x}{\left(\sin x+\cos x\right)^{2} } {\rm d}x\)

\(=\int _{0}^{\frac{\pi }{2} }\frac{1}{\sin x+\cos x} {\rm d}x=\frac{1}{\sqrt{2} } \int _{0}^{\frac{\pi }{2} }\frac{1}{\cos \left(x-\frac{\pi }{4} \right)} {\rm d}x    \)
\(\Rightarrow 2I=\frac{1}{\sqrt{2} } \int _{-\frac{\pi }{4} }^{\frac{\pi }{4} }\frac{1}{\cos u} {\rm d}u=\frac{1}{\sqrt{2} } \int _{-\frac{\pi }{4} }^{\frac{\pi }{4} }\frac{\cos u}{1-\sin ^{2} u} {\rm d}u\)

\(=\frac{1}{\sqrt{2} }\int _{-\frac{\pi }{4} }^{\frac{\pi }{4} }\frac{1}{\left(1-\sin x\right)\left(1+\sin x\right)} {\rm d}\left(\sin x\right)    \)
\(\Rightarrow 2I=\frac{1}{\sqrt{2} } .\frac{1}{2} \int _{-\frac{\pi }{4} }^{\frac{\pi }{4} }\left(\frac{1}{1-\sin x} -\frac{1}{1+\sin x} \right){\rm d}\left(\sin x\right) \)

\(=\frac{1}{2\sqrt{2} } \ln \left|\frac{\sin x+1}{\sin x-1} \right|\left|{}_{-\frac{\pi }{4} }^{\frac{\pi }{4} } =\right. \frac{\ln \left(17+12\sqrt{2} \right)}{2\sqrt{2} }  \)
Vậy \(I=\frac{\sqrt{2} \ln \left(17+12\sqrt{2} \right)}{8} .\)