0
$\begingroup$

I am struggling to solve this integral to the method of substitution/replacement method of variables. I know I have to find a u which the derivative of u (u') is present in the equation.

But I have trouble finding what will that my variable to replace . I have these two resolutions but of these I would not solve alone and the other I didn't understand anything.

My question is: Is there a simpler way to solve (even if it is a long resolution) where I can understand every step and make it more obvious which variable will replace?

$$ \int \frac{du}{u^2 + a^2}, (a\neq0) $$

Resolution 1 (I got the same steps but the difficulty is in understanding which replace variable) $$ \int \frac{du}{u^2 + a^2} = \frac{1}{a^2}\int\frac{du}{\frac{u^2}{a^2}+ 1} $$ $$ v = \frac{u}{a} \implies dv = \frac{1}{a}, du =adv $$ $$ \frac{1}{a^2}\int\frac{adv}{v^2+ 1} = \frac{1}{a}arctan(\frac{u}{a}) + C $$

Resolution 2: I have no idea what happened to the $+ a^2$ in $(atan\theta)^2 + a^2 = a^2sec^2\theta$ $$ \tan^2\theta +1 = sec^2\theta \\ a^2\tan^2\theta + a^2 = a^2\sec^2\theta\\ (a\tan\theta)^2 + a^2 = a^2\sec^2\theta\\ u = atg\theta, du = asec^2\theta d\theta\\ \int\frac{du}{u^2 + a^2} = \int\frac{a\sec^2\theta d\theta}{a^2\sec^2\theta} = \frac{1}{2}\int1d\theta = \frac{1}{a}\arctan(\frac{u}{a}) + C $$

$\endgroup$
3
  • 2
    $\begingroup$ suggestion: use backslash before standard functions. It will look a lot better. $\endgroup$ Oct 15, 2015 at 14:20
  • 1
    $\begingroup$ And \left and \right before parentheses (or any other pair of delimiters) to make them adjust to the size of the content. $\endgroup$
    – joriki
    Oct 15, 2015 at 14:25
  • $\begingroup$ the result is ok $\endgroup$ Oct 15, 2015 at 14:51

1 Answer 1

0
$\begingroup$

The method you have presented is basically a rehash of the fact that $$ \frac{d}{dx} \arctan(x) = \frac{1}{1 + x^2}$$ Note that $$\frac{d}{dx} \arctan(x/a) = \frac{1}{a}\frac{1}{1 + (x/a)^2} = \frac{a}{a^2 + x^2}$$ The above basically proves the result $$ \frac{1}{a^2 + x^2} = \frac{1}{a} \left( \frac{d}{dx} \arctan(a/x)\right)$$ Now it is "trivial" to see that $$\int \frac{dx}{a^2 + x^2} = \int \frac{1}{a} \left( \frac{d}{dx} \arctan(a/x)\right) dx = \frac{1}{a} \arctan(x/a) + C, \ C \in \mathbb{R}$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .