Mathematics

# The value of $\displaystyle\int\limits_{0}^{\frac{\pi}{4}} \tan^2 \theta\ d\theta=$

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

##### SOLUTION
Now,
$\displaystyle\int\limits_{0}^{\frac{\pi}{4}} \tan^2 \theta\ d\theta$
$=\displaystyle\int\limits_{0}^{\frac{\pi}{4}} (\sec^2 \theta-1)\ d\theta$ [ Since $\sec^2 \theta-\tan^2 \theta=1$]
$=\left[\tan \theta-\theta\right]_0^{\frac{\pi}{4}}$
$=1-\dfrac{\pi}{4}$.

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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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