Integration of a function within another function

sarvesh0303
Messages
61
Reaction score
2

Homework Statement



In differentiation we use the chain rule to find the derivative of a function within other function [f(g(x))]
Is there any similar method for indefinite integration for example how am I supposed to find the integral
of e^(4x^3+6x^2+8x+9) or sin(x^2).

Homework Equations


none


The Attempt at a Solution


Couldn't reach anywhere!
 
Physics news on Phys.org
<br /> \frac {df(g(x))} {dx} = f&#039;(g(x))g&#039;(x)<br /> \Rightarrow \int \frac {df(g(x))} {dx} dx = \int f&#039;(g(x))g&#039;(x) dx<br /> This is sometimes called "u-substitution", meaning one let's u = g(x), then <br /> \int f&#039;(g(x))g&#039;(x) dx = \int f&#039;(u) du = f(u) + C<br /> Finding a u-substitution is a matter of art (which means lots of practice and sometimes luck). But even that does not guarantee you can always take an indefinite analytically. The integrals you listed are of this kind. \int sin(x^2) dx is known as the Fresnel integral.
 
To add to voko's comment, since the u-substitution method is essentially the chain rule in reverse, if the function which you are trying to antidifferentiate did not come as the result of a chain rule calculation, you would not expect the u-substitution method to work. That is why, for example you can work$$
\int xe^{x^2}\, dx$$with the substitution ##u=x^2## but you can't work$$
\int e^{x^2}\, dx$$that way. You can tell that the first comes from a chain rule because, to within a constant, the integrand is the derivative of ##e^{x^2}## with the extra ##x## in the integral coming from the chain rule. That ##x## is missing in the second integral and it can not be worked with the u-substitution or, as it turns out, any other elementary method.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top