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Integration by substitution is an integration technique for definite integrals based upon this formula : $$\displaystyle\int_{a}^{b}f(\phi(x))\phi'(x)\,dx=\int_{\phi(a)}^{\phi(b)}{f(u)\,du}$$

I don't understand how it can be applied in the case of indefinite integrals, how it can be used to find antiderivatives, , as in this example : $$\displaystyle\int{\frac{dx}{x^2+a^2}}=\frac{1}{a}\int{\frac{du}{1+u^2}}=\frac{1}{a}\tan^{-1}(u)+C=\frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right)+C$$

The formula $\displaystyle\int_{a}^{b}{f(\phi(x))\phi'(x)\,dx}=\int_{\phi(a)}^{\phi(b)}{f(u)\,du}$ is for definite integrals. Why can it be applied to indefinite integrals ?

P.S. : I obviously already understand the algebra

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  • $\begingroup$ You may be interested in this. $\endgroup$ – David Mitra Feb 17 '13 at 21:05
  • $\begingroup$ There are two formulas, one for the indefinite integral, the other for the definite integral. The one for the indefinite integral is obtained by removing the "limits." There is of course a close connection between the two. $\endgroup$ – André Nicolas Feb 17 '13 at 21:06
  • $\begingroup$ What's the formula for the indefinite integral ? Where can I find it with a proof of it ? $\endgroup$ – user62705 Feb 17 '13 at 21:11
  • $\begingroup$ It is as I wrote, just remove the limits. In the notation of your post, let $u=\phi(x)$. Then $\int f(\phi(x))\phi'(x)\,dx=\int f(u)\,du$. The proof is the already proved Chain Rule for differentiation. $\endgroup$ – André Nicolas Feb 17 '13 at 21:23
  • $\begingroup$ @DavidMitra Nice document David. What is interesting is that it says that the equation $\int{f(\phi(x))\phi'(x)\,dx}=\int{f(u)\,du}$ is false ! Any comment about that ? $\endgroup$ – user62705 Feb 17 '13 at 22:07
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Let $F(u)$ be an antiderivative of $f(u)$. We show that $F(\phi(x))$ is an antiderivative of $f(\phi(x))\phi'(x)$.

The proof uses the Chain Rule. Differentiate $F(\phi(x))$. We get $\phi'(x)F'(\phi(x))$, which is $f(\phi(x))\phi'(x)$.

In your particular example, we want $\int \frac{dx}{x^2+a^2}$, where $a\ne 0$. We can rewrite this as $\int \frac{1}{a^2}\frac{dx}{1+(x/a)^2}$. Let $\phi(x)=x/a$. Then $\phi'(x)=1/a$. So we want $$\int \frac{1}{a}\frac{1}{1+(\phi(x))^2}\phi'(x)\,dx.$$ This is $\int \frac{1}{a} \int \frac{du}{1+u^2}$ where $u=\phi(x)$. We get $\frac{1}{a}\arctan(x/a)+C$.

One does not go through all of this writing when actually using substitution. Here is a medium length version of the same thing. Let $x=au$. Then $dx=a\,du$ and $a^2+x^2=a^2(1+u^2)$. Thus our integral is $$\int \frac{1}{a^2}\frac{1}{1+u^2}a\,du,$$ which is $\frac{1}{a}\arctan u+C$. Now replace $u$ by $x/a$.

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  • $\begingroup$ Thank you very much to all of you. $\endgroup$ – user62705 Feb 17 '13 at 21:58
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The example you gave can be done even without substitution, just take into account that if $\,f(x)\,$ is a derivable function, then we have

$$\int\frac{f'(x)}{1+f(x)^2}dx=\arctan f(x)+K$$

so

$$\int\frac{dx}{a^2+x^2}=\frac{1}{a}\int\frac{\frac{1}{a}dx}{1+\left(\frac{x}{a}\right)^2}=\frac{1}{a}\arctan\frac{x}{a}+K\ldots$$

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If you know that

$$\int_{a}^{b}{f(\phi(x))\phi^{'}(x)dx}=\int_{\phi(a)}^{\phi(b)}{f(u)du}$$

then Fundamental Theorem of Calculus tells you that this can also be used for antiderivatives. Indeed, an antiderivative of $f(\phi(x))\phi^{'}(x)$ is

$$\int_{a}^{x}{f(\phi(t))\phi^{'}(t)dt}=\int_{\phi(a)}^{\phi(x)}{f(u)du} \,.$$

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