The integral int_{7pi/4}^{7pi/3} sqrt{tan^2 x | vedclass

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Question

(A) Let $I = \int_{7\pi/4}^{7\pi/3} \sqrt{	an^2 x} \, dx$. Since $\sqrt{	an^2 x} = |	an x|$,we evaluate the integral in the interval $[7\pi/4, 7\pi

Vedclass · Accepted Answer

$\log(2\sqrt{2})$

Vedclass · Answer

(A) Let $I = \int_{7\pi/4}^{7\pi/3} \sqrt{	an^2 x} \, dx$. Since $\sqrt{	an^2 x} = |	an x|$,we evaluate the integral in the interval $[7\pi/4, 7\pi/3]$. In the interval $[7\pi/4, 2\pi]$,$	an x$ is negative,so $|	an x| = -	an x$. In the interval $[2\pi, 7\pi/3]$,$	an x$ is positive,so $|	an x| = 	an x$. Thus,$I = \int_{7\pi/4}^{2\pi} -	an x \, dx + \int_{2\pi}^{7\pi/3} 	an x \, dx$. $I = [\log|\cos x|]_{7\pi/4}^{2\pi} + [-\log|\cos x|]_{2\pi}^{7\pi/3}$. $I = (\log|\cos 2\pi| - \log|\cos(7\pi/4)|) - (\log|\cos(7\pi/3)| - \log|\cos 2\pi|)$. Since $\cos 2\pi = 1$,$\log 1 = 0$. $I = -\log|\cos(7\pi/4)| - \log|\cos(7\pi/3)| = -\log(1/\sqrt{2}) - \log(1/2)$. $I = \log(\sqrt{2}) + \log(2) = \log(\sqrt{2} 	imes 2) = \log(2\sqrt{2})$.

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

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