tích phân

Question

cho

$f(x)+f(-x)=\sqrt{2-2\cos 2x}$ .tính

$\int\limits_{\frac{-3\pi }{2}}^{\frac{3\pi }{2}}f(x)dx$

tran85295 · Answer 1 · 2/6/2018 1:24:01 PM

$I=\int_{\frac{-3\pi}2}^{\frac{3\pi}2}f(x){\rm d}x$

$J=\int_{\frac{-3\pi}2}^{\frac{3\pi}2}f(-x){\rm d}x$

Ta có

$\begin{align*}I+J&=\int_{\frac{-3\pi}2}^{\frac{3\pi}2}\sqrt{2-2\cos 2x}{\rm d}x\\ &=2\int_{\frac{-3\pi}2}^{\frac{3\pi}2}\sqrt{\sin^2x}{\rm d}x\\ &= 4\int_0^{\frac{3\pi}2}\left|\sin x\right|{\rm d}x \\ &=4\int_0^{\pi}\left|\sin x\right|{\rm d}x+4\int_{\pi}^{\frac{3\pi}2}\left|\sin x\right|{\rm d}x \\ &=4\int_0^{\pi}\sin x{\rm d}x+4\int_{\pi}^{\frac{3\pi}2}(-\sin x){\rm d}x \\ &=-4\cos x\bigg|_0^{\pi}+4\cos x\bigg|_{\pi}^{\frac{3\pi}2}\\&=12\end{align*}$

Mặt khác, khi đặt

$t=-x$ thì ta có:

$I=\int_{\frac{-3\pi}2}^{\frac{3\pi}2}f(x){\rm d}x=\int_{\frac{-3\pi}2}^{\frac{3\pi}2}f(t){\rm d}t=\int_{\frac{3\pi}2}^{\frac{-3\pi}2}f(-x){\rm d}\left(-x\right)=\int_{\frac{-3\pi}2}^{\frac{3\pi}2}f(-x){\rm d}x=J$

Suy ra

$I=J=6$ , vậy tích phân cần tìm là 6