Tính tích phân sau: ∫(1/(5^x/2+1)cos x/2)dx

Question

Tính tích phân sau:

\(\int _{\frac{-\pi }{2} }^{\frac{\pi }{2} }\frac{1}{\left(5^{\frac{x}{2} } +1\right)\cos \frac{x}{2} } dx\)

Answer 1

\(\int _{\frac{-\pi }{2} }^{\frac{\pi }{2} }\frac{1}{\left(5^{\frac{x}{2} } +1\right)\cos \frac{x}{2} } dx \)

 Đặt \(x=-t\Rightarrow dx=-dt\) .

Khi đó \(\int _{\frac{-\pi }{2} }^{\frac{\pi }{2} }\frac{1}{\left(5^{\frac{x}{2} } +1\right)\cos \frac{x}{2} } dx =\int _{\frac{\pi }{2} }^{-\frac{\pi }{2} }\frac{1}{\left(5^{\frac{-t}{2} } +1\right)\cos \frac{-t}{2} } \left(-dt\right)\)

\( =\int _{\frac{-\pi }{2} }^{\frac{\pi }{2} }\frac{5^{\frac{t}{2} } }{\left(5^{\frac{t}{2} } +1\right)\cos \frac{t}{2} } dt =\int _{\frac{-\pi }{2} }^{\frac{\pi }{2} }\frac{5^{\frac{x}{2} } }{\left(5^{\frac{x}{2} } +1\right)\cos \frac{x}{2} } dx \)
\(\Rightarrow \int _{\frac{-\pi }{2} }^{\frac{\pi }{2} }\frac{1}{\left(5^{\frac{x}{2} } +1\right)\cos \frac{x}{2} } dx +\int _{\frac{-\pi }{2} }^{\frac{\pi }{2} }\frac{5^{\frac{x}{2} } }{\left(5^{\frac{x}{2} } +1\right)\cos \frac{x}{2} } dx\)

\(=\int _{\frac{-\pi }{2} }^{\frac{\pi }{2} }\frac{1}{\cos \frac{x}{2} } dx  \)
Ta có: \(\int _{\frac{-\pi }{2} }^{\frac{\pi }{2} }\frac{1}{\cos \frac{x}{2} } dx =2\int _{\frac{-\pi }{2} }^{\frac{\pi }{2} }\frac{1}{\cos \frac{x}{2} } d\left(\frac{x}{2} \right)  \)

\(=2\int _{\frac{-\pi }{2} }^{\frac{\pi }{2} }\frac{\cos \frac{x}{2} }{\cos ^{2} \frac{x}{2} } d\left(\frac{x}{2} \right) =2\int _{\frac{-\pi }{2} }^{\frac{\pi }{2} }\frac{1}{\left(1+\sin \frac{x}{2} \right)\left(1-\sin \frac{x}{2} \right)} d\left(\sin \frac{x}{2} \right)\)
\(=\ln \left|\frac{\sin \frac{x}{2} +1}{\sin \frac{x}{2} -1} \right|\left|{}_{-\frac{\pi }{2} }^{\frac{\pi }{2} } \right. =2\ln \frac{2+\sqrt{2} }{2-\sqrt{2} }  \)

\(  \Rightarrow \int _{\frac{-\pi }{2} }^{\frac{\pi }{2} }\frac{1}{\left(5^{\frac{x}{2} } +1\right)\cos \frac{x}{2} } dx =\ln \frac{2+\sqrt{2} }{2-\sqrt{2} }  . \)