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When using trigonometric substitution to find the indefinite integral of an expression, the derivative typically begins as $dx$. Once some expression is substituted for the $x$ in the expression, the derivative of the function is also changed. Why is that?

Examples of trigonometric substitution are here.

Community
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Kelmikra
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2 Answers2

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It's just the chain rule: $$ \int \underbrace{f'(g(x))}\ \underbrace{g'(x)\,dx} = \int f'(u)\,du = f(u)+C = f(g(x))+C $$

Postscript in response to comments:

The chain rule is differentiation by substitution. Suppose $y = f(g(x))$. Then do the substitution $u=g(x)$. Then $y=f(u)$ and $$\frac{dy}{dx}=(f\circ g)'(x)=f'(u)g'(x).$$ Notice that one can write $$dy = f'(u)\ \underbrace{g'(x)\,dx} = f'(u)\,du.$$

That's the chain rule and that's differentiation by substitution.

Michael Hardy
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  • I don't see how this is the chain rule. As far as I know, the chain rule doesn't use substitutions. – Kelmikra Jan 23 '15 at 02:14
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    @Kyth'Py1k The chain rule says that $[f(g(x))]'=f'(g(x))g'(x)$. Then, $\int f'(g(x))g'(x) ; dx = \int [f(g(x))]' ; dx$. Can you conclude? – Gahawar Jan 23 '15 at 02:21
  • @Kyth'Py1k The chain rule doesn't use substitutions; but this substitution uses the chain rule. – Graham Kemp Jan 23 '15 at 03:21
  • @Gahawar The chain rule doesn't change $dx$, but trigonometric substitution does. That's what I really don't understand. I don't even understand what $dx$ really is. It's not a real number, it's infinitely small, and yet some derivatives are larger than others. – Kelmikra Jan 23 '15 at 13:12
  • @Kyth'Py1k See this thread for more information on $dx$ (coincidentally, the answer by Michael Hardy is very good): http://math.stackexchange.com/questions/200393/what-is-dx-in-integration – Gahawar Jan 23 '15 at 13:30
  • @Gahawar Thanks for the link; I think I understand what $dx$ means now. I still don't know why trigonometric substitution changes $dx$ while the standard use of the chain rule doesn't though. – Kelmikra Jan 23 '15 at 17:20
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    The chain rule is differentiation by substitution. Suppose $y = f(g(x))$. Then do the substitution $u=g(x)$. Then $y=f(u)$ and $\dfrac{dy}{dx}=(f\circ g)'(x)=f'(u)g'(x)$. Notice that one can write $dy = f'(u)\ \underbrace{g'(x),dx}$, and then since $du=g'(x),dx$, we have $dy = f'(u),du$. That's the chain rule and that's differentiation by substitution. ${}\qquad{}$ – Michael Hardy Jan 24 '15 at 00:29
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when you use a transformation say $(x, f(x))$ to $(u, f(u)$ where $u = u(x)$ the domain of $f(u)$ is changed.

for example let us take $f(x) = (2x)^3$ on $D = [0,1]$ and consider the transformation $x = u/2.$ the transformed function $f(x) = g(u) = u^3$ and the domain of $g$ is $[0,2].$

if you are finding the area $\int_0^1 (2x)^3 \ dx = 2$ which is the same as $\int_0^2 u^3 \frac{du}{2} = 2$

you can think of the transformation sends the rectangle $f(x)dx$ to $f(u/2) d(u/2)$

i mean by transformation of rectangles is this: draw a graph $y = 8x^3$ on $0 \le x \le 1.$ partition the interval $[0,1]$ into ten equal pieces. pick a piece say $[3/10, 4/10]$ that is the base of the rectangle. now you pick any point $c$ in this base and make a rectangle of height $f(c)$ by the transformation $u = 2x,$ we mean all the $x$ values are multiplied by two(magnified/stretched) and the $y$ values stay the same. the base $[3/10, 4/10]$ in the $xy$ will go to $6/10,8/10]$ the height of the rectangle stays put. to take care of the stretch we needed to divide by $2$ that is why you see that $du/2.$

i hope this helps. it is a good idea and to draw the graphs of $y = 8x^3$ and do a transformation $u = 2x.$

abel
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  • How are the two equations on the second to last line the same? One equals two and the other equals four. Could you also explain what you mean by "the transformation send the rectangle"? – Kelmikra Jan 23 '15 at 02:12