Mathematics

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


ANSWER

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


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
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