Mathematics

# Evaluate the following definite integrals :$\displaystyle \int _{0}^1 \dfrac {1}{1+x^2}dx$

##### SOLUTION
Let $x=\tan{\theta}\Rightarrow\,dx={\sec}^{2}{\theta}d\theta$

When $x=0\Rightarrow\,\theta=0$

When $x=1\Rightarrow\,\theta=\dfrac{\pi}{4}$

$\displaystyle\int_{0}^{1}{\dfrac{dx}{1+{x}^{2}}}$

$=\displaystyle\int_{0}^{\frac{\pi}{4}}{\dfrac{{\sec}^{2}{\theta}d\theta}{1+{\tan}^{2}{\theta}}}$

$=\displaystyle\int_{0}^{\frac{\pi}{4}}{\dfrac{{\sec}^{2}{\theta}d\theta}{{\sec}^{2}{\theta}}}$

$=\displaystyle\int_{0}^{\frac{\pi}{4}}{d\theta}$

$=\left[\theta\right]_{0}^{\frac{\pi}{4}}$

$=\dfrac{\pi}{4}-0=\dfrac{\pi}{4}$

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

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