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

Question

Tính tích phân sau:

\(I = \int _{0}^{\frac{\pi }{4} }\frac{xcosx+\left(x-1\right)\sin x}{1+\tan x}  {\rm d}x \)

Answer 1

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

\(=\int _{0}^{\frac{\pi }{4} }x\cos x{\rm dd}x -\int _{0}^{\frac{\pi }{4} }\frac{\sin x\cos x}{\cos x+\sin x}  {\rm d}x=I_{1} -I_{2} \) 

 Tính \(I_{1} =\int _{0}^{\frac{\pi }{4} }x\cos x{\rm d}x\)  

 Đặt \(\left\{\begin{array}{l} {u=x} \\ {dv=\cos xdx} \end{array}\right. \Rightarrow \left\{\begin{array}{l} {du=dx} \\ {v=\sin x} \end{array}\right.  \)
\(I_{1} =x\sin x|_{0}^{\frac{\pi }{4} } -\int _{0}^{\frac{\pi }{4} }\sin xdx =\left(x\sin x+\cos x\right)|_{0}^{\frac{\pi }{4} }   \)

\(=\left(\frac{\pi }{4} .\frac{\sqrt{2} }{2} +\frac{\sqrt{2} }{2} \right)-\left(0+1\right)=\frac{\sqrt{2} \pi }{8} +\frac{\sqrt{2} }{2} -1 \)

 + Tính \( I_{2} =\int _{0}^{\frac{\pi }{4} }\frac{\sin 2x}{2\sqrt{2} \sin \left(x+\frac{\pi }{4} \right)}  {\rm d}x.\) 

 Đặt \(t=x+\frac{\pi }{4}\)  thì \(dt=dx \)
\(I_{2} =\int _{\frac{\pi }{4} }^{\frac{\pi }{2} }\frac{-cos2t}{2\sqrt{2} \sin t} dt=\int _{\frac{\pi }{4} }^{\frac{\pi }{2} }\frac{1}{\sqrt{2} } \sin tdt-\int _{\frac{\pi }{4} }^{\frac{\pi }{2} }\frac{1}{2\sqrt{2} \sin t} dt \) 

\(=\left(-\frac{1}{\sqrt{2} } \cos x\right)|_{\frac{\pi }{4} }^{\frac{\pi }{2} } -\frac{1}{2\sqrt{2} } \ln \left|\tan \frac{x}{2} \right||_{\frac{\pi }{4} }^{\frac{\pi }{2} }  \)
\(=\frac{1}{2} +\frac{1}{2\sqrt{2} } \ln \left|\tan \frac{\pi }{8} \right|.\) 

 Suy ra \(I=I_{1} -I_{2} =\frac{\sqrt{2} \pi }{8} +\frac{\sqrt{2} }{2} -\frac{3}{2} -\frac{1}{2\sqrt{2} } \ln \left|\tan \frac{\pi }{8} \right|\)