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
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