Mathematics

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

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

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

When $x=-1\Rightarrow\,\tan{\theta}=-1\Rightarrow\,\theta={\tan}^{-1}{\left(-1\right)}=\dfrac{-\pi}{4}$

We know that $\displaystyle\int_{-a}^{a}{f\left(x\right)dx}=2\displaystyle\int_{0}^{a}{f\left(x\right)dx}$

$\Rightarrow\,\displaystyle\int_{-1}^{1}{\dfrac{dx}{1+{x}^{2}}}$

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

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

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

$=2\left[\theta\right]_{0}^{\frac{\pi}{4}}$

$=2\left[\dfrac{\pi}{4}-0\right]=\dfrac{\pi}{2}$

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