Trigonometric Substitution: Problem 4

Solve:       $latex A = \int  \frac {1}{(4+x^2)^2}  dx$

This is the form of arctan so we use the substitution

$2tan(\theta) = x

the “2” is from “4” + x^2

dx = 2  sec^2\theta d\theta

Simplify: 4+x^2 = 4(1+tan^2\theta) = 4 sec^2\theta

Substitute: A = \int \frac{2  sec^2\theta d\theta}{(4sec^2\theta)^2}

= 1/8 \int cos^2\theta d\theta

We have solved \int cos^2\theta d\theta    \qquad before.

If you work through it you should get

\int cos^2\theta d\theta = \frac{\theta}{2} + \frac{2sin\theta cos\theta}{4}

Substitute for \theta : \theta = arctan(x/2)

sin\theta = \frac{x}{\sqrt{x^2 +4}}

and    \qquad cos\theta = \frac{{2}}{\sqrt{x^2 +4}}

Prove out the final result:$

$A = \frac{ arctan( x/2)}{2} +\frac {{x}}{x^2 +4}  + C

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