Mathematics

Evaluate the following definite integral:

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


SOLUTION
Let $$I=\displaystyle \int _0^1 \dfrac {1-x^2}{(1+x^2)^2} dx$$

let $$x=\tan \theta$$

$$dx=\sec^2 \theta d\theta$$

$$I=\displaystyle \int _0^{\pi /4}\dfrac {1-\tan^2 \theta}{1+\tan^2 \theta}d\theta =\displaystyle \int _0^{\pi /4} \cos^2 \theta \ d\theta$$

$$I=\dfrac {\sin 2\theta}{2} |_0^{\pi /4}=\dfrac {1}{2}$$

Hence $$I=\dfrac {1}{2}$$
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Subjective Medium Published on 17th 09, 2020
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