MATH105 April 2018
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Question 02 (b)

Use the substitution $x=\tan \theta$ to compute $\int _{0}^{1}{\frac {x^{2}}{(1+x^{2})^{2}}}dx$.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint

Use trigonometric identities $1+\tan ^{2}\theta =\sec ^{2}\theta$ and $\sin ^{2}{\frac {u}{2}}={\frac {1\cos u}{2}}$.

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Solution

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As suggested in the question, we use the substitution $x=\tan \theta$. Then, we have $dx=\sec ^{2}\theta d\theta$. Recall the trigonometric identity $1+\tan ^{2}\theta =\sec ^{2}\theta$. Then, together with $\tan 0=0$ and $\tan {\frac {\pi }{4}}=1$, the integral becomes
${\begin{aligned}\int _{0}^{1}{\frac {x^{2}}{(1+x^{2})^{2}}}dx&=\int _{0}^{\frac {\pi }{4}}{\frac {\tan ^{2}\theta }{(1+\tan ^{2}\theta )^{2}}}\sec ^{2}\theta d\theta \\&=\int _{0}^{\frac {\pi }{4}}{\frac {\tan ^{2}\theta }{\sec ^{4}\theta }}\sec ^{2}\theta d\theta =\int _{0}^{\frac {\pi }{4}}{\frac {\tan ^{2}\theta }{\sec ^{2}\theta }}d\theta \\&=\int _{0}^{\frac {\pi }{4}}\sin ^{2}\theta d\theta .\end{aligned}}$
Then, we apply the trigonometric identity $\sin ^{2}{\frac {u}{2}}={\frac {1\cos u}{2}}$ for $u=2\theta$, we can evaluate the integral:
$\int _{0}^{\frac {\pi }{4}}\sin ^{2}\theta d\theta =\int _{0}^{\frac {\pi }{4}}{\frac {1\cos 2\theta }{2}}d\theta =\left[{\frac {1}{2}}\theta {\frac {1}{4}}\sin 2\theta \right]_{0}^{\frac {\pi }{4}}={\frac {1}{2}}\cdot {\frac {\pi }{4}}{\frac {1}{4}}\left(\sin {\frac {\pi }{2}}\sin 0\right)={\frac {\pi }{8}}{\frac {1}{4}}$
To summarize, we get
$\int _{0}^{1}{\frac {x^{2}}{(1+x^{2})^{2}}}dx={\frac {\pi }{8}}{\frac {1}{4}}$
Answer: $\color {blue}{\frac {\pi }{8}}{\frac {1}{4}}$
