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
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

#### Realted Questions

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Q2 Subjective Medium
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Q4 Single Correct Hard
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