The integral _7/4^7/3 ^2x,dx is equal to:

Question

The integral $\displaystyle\int_{7\pi/4}^{7\pi/3} \sqrt{	an^2x}\,dx$ is equal to:

Accepted Answer

Let $I = \int_{7\pi/4}^{7\pi/3} 	an x\,dx = -\log|\cos x|\Big|_{7\pi/4}^{7\pi/3}$

$= \log\dfrac{\cos(7\pi/4)}{\cos(7\pi/3)} = \log\dfrac{\cos(2\pi-\pi/4)}{\cos(2\pi+\pi/3)} = \log\dfrac{\cos(\pi/4)}{\cos(\pi/3)} = \log\dfrac{1/\sqrt2}{1/2} = \log\sqrt2$

Solutions · 1