Tính các tích phân sau: ∫(x^3−3x+1)/(cos^2 x)dx

Question

Tính các tích phân sau:

\(\int _{\frac{-\pi }{4} }^{\frac{\pi }{4} }\frac{x^{3} -3x+1}{c{\rm os}^{2} x} \, \,  {\rm d}x \)

Answer 1

\(I_{3} =\int _{\frac{-\pi }{4} }^{\frac{\pi }{4} }\frac{x^{3} -3x+1}{c{\rm os}^{2} x} \, \,  dx\)

 Ta có
\(I =\int _{\frac{-\pi }{4} }^{\frac{\pi }{4} }\frac{x^{3} -3x}{\cos ^{2} x} \, \,  dx+\int _{\frac{-\pi }{4} }^{\frac{\pi }{4} }\frac{1}{\cos ^{2} x} \, \,  dx\)

\(=\int _{\frac{-\pi }{4} }^{\frac{\pi }{4} }\frac{x^{3} -3x}{\cos ^{2} x} \, \,  dx+\tan x\left|\begin{array}{l} {\frac{\pi }{4} } \\ {\frac{-\pi }{4} } \end{array}\right. =\int _{\frac{-\pi }{4} }^{\frac{\pi }{4} }\frac{x^{3} -3x}{\cos ^{2} x} \, \,  dx+2\)
Xét  \(J=\int _{\frac{-\pi }{4} }^{\frac{\pi }{4} }\frac{x^{3} -3x}{c{\rm os}^{2} x} \, \,  dx      \)

 Đặt  \(x=-t\Rightarrow dx=-dt\), khi đó :
\(J=-\int _{\frac{\pi }{4} }^{\frac{-\pi }{4} }\frac{(-t)^{3} -3(-t)}{c{\rm os}^{2} (-t)} \, \,  dt=\int _{\frac{-\pi }{4} }^{\frac{\pi }{4} }\frac{-t^{3} +3t}{c{\rm os}^{2} t} \, \,  dt\)

\(=\int _{\frac{-\pi }{4} }^{\frac{\pi }{4} }\frac{-x^{3} +3x}{c{\rm os}^{2} x} \, \,  dx=-J \)
Suy ra :    \(2J=0\Rightarrow J=0\)

 Vậy:   \(I =J+2=2.\)