Tính tích phân sau: ∫(cos^4 x+sin^4 x)/(2^x+1)dx

Question

Tính tích phân sau:

\(\int _{-\frac{\pi }{2} }^{\frac{\pi }{2} }\frac{\cos ^{4} x+\sin ^{4} x}{2^{x} +1}  dx \)

Answer 1

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

\(=\frac{3}{4} \int _{-\frac{\pi }{2} }^{\frac{\pi }{2} }\frac{1}{2^{x} +1}  dx+\frac{1}{4} \int _{-\frac{\pi }{2} }^{\frac{\pi }{2} }\frac{\cos 4x}{2^{x} +1}  dx. \)
Xét\( I_{2}^{} =\int _{-\frac{\pi }{2} }^{\frac{\pi }{2} }\frac{\cos 4x}{2^{x} +1}  dx.\)

\(Đặt t=-x\Rightarrow dx=-dt.\)
\(I_{1}^{} =\int _{-\frac{\pi }{2} }^{0}\frac{\cos 4x}{2^{x} +1}  dx+\int _{0}^{\frac{\pi }{2} }\frac{\cos 4x}{2^{x} +1}  dx  (*). \)
\(\Rightarrow I_{1a}^{} =\int _{-\frac{\pi }{2} }^{0}\frac{\cos 2x}{3^{x} +1}  dx=-\int _{\frac{\pi }{2} }^{0}\frac{\cos 4t}{2^{-t} +1}  dt\)

\(=\int _{0}^{\frac{\pi }{2} }\frac{2^{t} .\cos 4t}{2^{t} +1}  dt=\int _{0}^{\frac{\pi }{2} }\frac{2^{x} .\cos 4x}{2^{x} +1}  dx. \)
Thay vào (*) ta được:

\(I_{1}^{} =\int _{0}^{\frac{\pi }{2} }\cos 4x dx=\frac{1}{4} \sin 4x\left|\begin{array}{l} {\frac{\pi }{2} } \\ {0} \end{array}\right. =0.\)

 Xét \(I_{2}^{} =\frac{3}{4} \int _{-\frac{\pi }{2} }^{\frac{\pi }{2} }\frac{1}{2^{x} +1}  dx\).

\(=\frac{3}{4} \int _{-\frac{\pi }{2} }^{0}\frac{1}{2^{x} +1}  dx+\frac{3}{4} \int _{0}^{\frac{\pi }{2} }\frac{1}{2^{x} +1}  dx. (2*)\)

 Đặt \(t=-x\Rightarrow dx=-dt. \)

\(\Rightarrow I_{2a}^{} =-\frac{3}{4} \int _{\frac{\pi }{2} }^{0}\frac{1}{2^{-t} +1}  dt=\frac{3}{4} \int _{0}^{\frac{\pi }{2} }\frac{2^{t} }{2^{t} +1}  dt.\)

 Thay vào (2*) ta được: \(I_{2}^{} =\frac{3}{4} \int _{0}^{\frac{\pi }{2} }\frac{2^{x} }{2^{x} +1}  dx+\frac{3}{4} \int _{0}^{\frac{\pi }{2} }\frac{1}{2^{x} +1}  dx\).    

\(=\frac{3}{4} \int _{0}^{\frac{\pi }{2} }dx= \frac{3}{4} x\left|\begin{array}{l} {\frac{\pi }{2} } \\ {0} \end{array}\right. =\frac{3\pi }{8} \)

  Vậy \(I=\frac{3\pi }{8} .   \)