Mathematics

# The value of $\int _{ \frac { 7\pi }{ 4 } }^{ \frac { 7\pi }{ 3 } }{ \sqrt { { \tan }^{ 2 }x } dx }$ is equal to

$\log \sqrt 2$

##### SOLUTION
$s = \int_{\frac{7\pi }{4}}^{\frac{7\pi}{3}} \sqrt{tan^{2}x} dx = \int_{\frac{7\pi}{4}}^{\frac{7\pi}{3}}tan xdx.$
$=\int_{\frac{7\pi}{4}}^{\frac{7\pi}{3}} \frac{sin x dx}{cosx}$
Let cos x = t.                             At $x = \frac{7\pi}{4} , t = cos\frac{7\pi}{4}$
- sinxdx = dt.                                   $= cos (2\pi-\frac{\pi}{4}) = cos\frac{\pi}{4} =\frac{1}{2}$
$\Rightarrow sin xdx = - dt A t x = \frac{7\pi }{3}, t= cos\frac{7\pi }{3}= cos (\pi +\frac{\pi }{3}) =+cos \frac{\pi }{3} =\frac{1}{2}$
$x = \int_{\frac{1}{\sqrt{2}}}^{\frac{1}{2}}\frac{-dt}{t}$
$= \left [ -log t \right ]^{\frac{1}{2}}_{\frac{1}{\sqrt{2}}}$
$= -log \frac{1}{2}+log\frac{1}{\sqrt{\frac{1}{2}}}$
$= log\frac{1}{\sqrt{2}}-log\frac{1}{2}$
$= log \frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}}$
$= log \frac{2}{\sqrt{2}}$
$= log \frac{2\sqrt{2}}{2}$
$= log \sqrt{2}$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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