Key to "Proving an integral∫dxa2+x2=1aarctan⁡xa+Cwhere C is constant."

nastupnat0hh

Answered question

2022-03-31

Proving an integral

\int \frac{d x}{a^{2} + x^{2}} = \frac{1}{a} \arctan \frac{x}{a} + C

where C is constant.

Jambrichp2w2

Beginner2022-04-01Added 12 answers

A better way to prove it would be to let

x = a \tan θ

, then

\frac{d x}{d θ} = a \sec^{2} θ \Rightarrow d x = a \sec^{2} θ d θ

. So:

\int \frac{1}{x^{2} + a^{2}} d x = \int \frac{a \sec^{2} θ}{a^{2} \tan^{2} θ + a^{2}} d θ

= \frac{1}{a} \int \frac{a \sec^{2} θ}{a (\tan^{2} θ + 1)} d θ

= \frac{1}{a} \int \frac{a \sec^{2} θ}{a \sec^{2} θ}

= \frac{1}{a} \int d θ

= \frac{1}{a} θ + k

= \frac{1}{a} \arctan \frac{x}{a} + k

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