A) \[\ell n\,2\]
B) \[\ell n\,4\]
C) \[\frac{1}{2}\ell n\,2\]
D) \[\frac{3}{2}\ell n\,2\]
Correct Answer: D
Solution :
Let \[I=\int\limits_{\frac{7\pi }{4}}^{\frac{7\pi }{3}}{\sqrt{{{\tan }^{2}}x\,dx\,=}}\int\limits_{\frac{7\pi }{4}}^{\frac{7\pi }{3}}{|\,\tan \,x|\,dx}\] \[=\int\limits_{\frac{7\pi }{4}}^{2\pi }{-(\tan \,x)dx+}\int\limits_{2\pi }^{\frac{7\pi }{3}}{(\tan \,x)dx}\] \[=(\ell n(\cos \,x))_{315{}^\circ }^{360{}^\circ }-(\ell n(\cos x))_{360{}^\circ }^{420{}^\circ }\] \[=\left( 0-\ell n\frac{1}{\sqrt{2}} \right)-\left( \ell n\frac{1}{2}-0 \right)\] \[=\frac{-1}{2}\ell n\frac{1}{2}-\ell n\frac{1}{2}=\frac{-3}{2}\ell n\frac{1}{2}=\frac{3}{2}\ell n\,\,2\]
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