Tính tích phân sau: I=∫((1+xcosx)sin2x)/(1+sinx)dx

Question

Tính tích phân sau: 

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

Answer 1

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

\(=\int _{0}^{\frac{\pi }{2} }\frac{2\sin xcosx}{1+\sin x} {\rm d}x +\int _{0}^{\frac{\pi }{2} }2x\sin x\left(1-\sin x\right){\rm d}x\)  
\(=\int _{0}^{\frac{\pi }{2} }\frac{2\sin xcosx}{1+\sin x} {\rm d}x +\int _{0}^{\frac{\pi }{2} }2x\sin x{\rm d}x -\int _{0}^{\frac{\pi }{2} }2x.\frac{1-cos2x}{2} {\rm d}x  \)
\(=\int _{0}^{\frac{\pi }{2} }\frac{2\sin xcosx}{1+\sin x} {\rm d}x+\int _{0}^{\frac{\pi }{2} }2x\sin x{\rm d}x  -\int _{0}^{\frac{\pi }{2} }xdx +\int _{0}^{\frac{\pi }{2} }xcos2x{\rm d}x  \)
+ Tính \( I_{1} =\int _{0}^{\frac{\pi }{2} }\frac{2\sin xcosx}{1+\sin x} dx \).

Đặt \(t=1+\sin x \Rightarrow dt=\cos xdx \)
\(I_{1} =\int _{1}^{2}\frac{2\left(t-1\right)}{t} dt =2\left(t-\ln \left|t\right|\right)|_{1}^{2} =2\left(2-\ln 2\right)-2=2-2\ln 2 \)
+ Tính \(x I_{2} =\int _{0}^{\frac{\pi }{2} }2x\sin xdx .\)

\(Đặt \left\{\begin{array}{l} {u=2x} \\ {dv=\sin xdx} \end{array}\right. \Rightarrow \left\{\begin{array}{l} {du=2dx} \\ {v=-cosx} \end{array}\right.  \)
\(I_{2} =-2xcosx|_{0}^{\frac{\pi }{2} } +\int _{0}^{\frac{\pi }{2} }2\cos x{\rm d}x =\left(-2xcosx+2\sin x\right)|_{0}^{\frac{\pi }{2} } =2 \)
+ Tính \(I_{3} =\int _{0}^{\frac{\pi }{2} }xcos2x{\rm d}x\) .

\(Đặt \left\{\begin{array}{l} {u=x} \\ {{\rm d}v=cos2x{\rm d}x} \end{array}\right. \Rightarrow \left\{\begin{array}{l} {{\rm d}u={\rm d}x} \\ {v=\frac{1}{2} \sin 2x} \end{array}\right.  \)
\(I_{3} =\frac{1}{2} x\sin 2x|_{0}^{\frac{\pi }{2} } -\int _{0}^{\frac{\pi }{2} }\frac{1}{2} \sin 2x{\rm d}x \) 

\(=\left(\frac{1}{2} x\sin 2x+\frac{1}{4} cos2x\right)|_{0}^{\frac{\pi }{2} } =-\frac{1}{4} -\frac{1}{4} =-\frac{1}{2} \)
Vậy \(I=\left(2-2\ln 2\right)+2-\frac{\pi ^{2} }{8} -\frac{1}{2} =\frac{7}{2} -\frac{\pi ^{2} }{8} -2\ln 2.\)