Mathematics

# $\int_{0}^{\pi/4}\tan^{2}\ x\ dx$

$1-\dfrac {\pi}{4}$

##### SOLUTION
$\displaystyle\int_{0}^{\frac{\pi}{4}}{{\tan}^{2}{x}dx}$
$=\displaystyle\int_{0}^{\frac{\pi}{4}}{\left({\sec}^{2}{x}-1\right)dx}$
$=\left[\tan{x}-x\right]_{0}^{\frac{\pi}{4}}$
$=\left[\left(\tan{\dfrac{\pi}{4}}-0\right)-\left(\dfrac{\pi}{4}-0\right)\right]$
$=\left[\left(1-0\right)-\left(\dfrac{\pi}{4}-0\right)\right]$
$=1-\dfrac{\pi}{4}$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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