Tính tích phân sau: ∫[ln(x+√x^2+1)+sin x.sin 3x]dx

Question

Tính tích phân sau:

\(\int _{-\frac{\pi }{2} }^{\frac{\pi }{2} }\left[\ln \left(x+\sqrt{x^{2} +1} \right)+\sin x.\sin 3x\right]{\rm d}x  \)

Answer 1

Đặt \(I=\int _{-\frac{\pi }{2} }^{\frac{\pi }{2} }\left[\ln \left(x+\sqrt{x^{2} +1} \right)+\sin x.\sin 3x\right]dx \)
\(=\int _{-\frac{\pi }{2} }^{\frac{\pi }{2} }\ln \left(x+\sqrt{x^{2} +1} \right)dx +\int _{-\frac{\pi }{2} }^{\frac{\pi }{2} }\sin x.\sin 3x.dx =I_{1} +I_{2} \) 
\(I_{1} =\int _{\frac{-\pi }{2} }^{\frac{\pi }{2} }\ln (x+\sqrt{x^{2} +1} ) \, dx \)
Đặt  \(x=-t\Rightarrow dx=-dt\), khi đó :
\(I_{1} =\int _{\frac{-\pi }{2} }^{\frac{\pi }{2} }\ln (x+\sqrt{x^{2} +1} ) \, dx=-\int _{\frac{\pi }{2} }^{\frac{-\pi }{2} }\ln (-x+\sqrt{x^{2} +1} \, ) \, dx\) 

\(=\int _{\frac{-\pi }{2} }^{\frac{\pi }{2} }\ln (-x+\sqrt{x^{2} +1} \, ) \, dx=\int _{\frac{-\pi }{2} }^{\frac{\pi }{2} }\ln \frac{1}{x+\sqrt{x^{2} +1} }  \, dx\)
\(=-\int _{\frac{-\pi }{2} }^{\frac{\pi }{2} }\ln (x+\sqrt{x^{2} +1} ) \, dx=-I_{1} . \)
Suy ra :    \(2I_{1} =0\Rightarrow I_{1} =0\)

\( I_{2} =\int _{-\frac{\pi }{2} }^{\frac{\pi }{2} }\sin x.\sin 3x.dx =\frac{1}{2} \int _{-\frac{\pi }{2} }^{\frac{\pi }{2} }\left(\cos 2x-\cos 4x\right).dx​​​​​​​\)
\(=\frac{1}{2} \int _{-\frac{\pi }{2} }^{\frac{\pi }{2} }\cos 2x.dx -\frac{1}{2} \int _{-\frac{\pi }{2} }^{\frac{\pi }{2} }\cos 4x.dx \)

\( =\frac{1}{4} \sin 2x\left|\begin{array}{l} {\frac{\pi }{2} } \\ {-\frac{\pi }{2} } \end{array}\right. -\frac{1}{8} \sin 4x\left|\begin{array}{l} {\frac{\pi }{2} } \\ {-\frac{\pi }{2} } \end{array}\right. \) 

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

Vậy \(I=0\)