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
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