Mathematics

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


ANSWER

$$\log \sqrt 2$$


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
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