The substitution method is a powerful tool in our efforts to compute integrals and antiderivatives. It is essentially the inverse to the chain rule.
Substitution Method: Indefinite Integrals
If \(F'(x)=f(x)\), i.e. if \(F\) is an antiderivative for \(f\), then: \[\int f(u(x))u'(x)dx = F(u(x)) + C\] Note that the RHS can also be thought of as \(\int f(u)du\) with an understanding the at the end we would substitute \(u=u(x)\). With that in mind, the substitution method is also written: \[\int f(u(x))u'(x)dx = \int f(u)du\]
This follows from direct observation: The derivative of \(F(u(x))\) would equal \(F'(u(x))u'(x) = f(u(x))u'(x)\).
Example: Suppose we had to compute \(\int x\sin(x^2)dx\). We can then consider \(u(x) = x^2\), and we would have: \[\int x\sin(x^2)dx = \frac{1}{2}\int \sin(x^2)(2x)dx = \frac{1}{2}\int\sin u du = -\frac{1}{2}\cos u = -\frac{1}{2}\cos(x^2) + C\]
In general to carry out the substitution:
As another example, consider the following: \[\int \frac{x^3}{(1+x^2)^3}dx\] Here the denominator is problematic, and so it is a good candidate for a substitution: \[u = 1+x^2\] We then need to compute: \[du = (1+x^2)'dx = 2xdx\] This leaves an \(x^2\) in the numerator, and we need to replace it with a suitable expression of \(u\): \[x^2 = u-1\] Finally, our integral becomes: \[\int \frac{u-1}{u^3}\frac{du}{2}\] We can then compute this integral by braking it up in two pieces: \[\int \frac{u-1}{u^3}\frac{du}{2} = \frac{1}{2}\int \frac{u}{u^3}du - \frac{1}{2}\int \frac{1}{u^3}du = \frac{1}{2}\int \frac{1}{u^2}du - \frac{1}{2}\int \frac{1}{u^3}du = - \frac{1}{2u} + \frac{1}{4u^2} = \frac{-2u + 1}{4u^2} + C\] Finally, we put back in \(u=1+x^2\): \[\int \frac{x^3}{(1+x^2)^3}dx = \frac{-2(1+x^2) + 1}{4(1+x^2)^2}\]
There if a version of the substitution method for definite integrals. The main difference is that the endpoints change:
Substitution Method: Definite Integrals
\[\int_a^b f(u(x))u'(x)dx = \int_{u(a)}^{u(b)}f(u)du\]
In other words:
As an example, let us compute the integral: \[\int_0^{\frac{\pi}{2}}\cos x\sin xdx\] We can do here a substitution \(u=\sin x\). Then \(du=\cos x dx\). The endpoints will change: When \(x=0\) we have \(u=\sin 0 = 0\) and when \(x=\frac{pi}{2}\) we have \(u=\sin \frac{pi}{2} = 1\). So we get the integral: \[\int_0^{\frac{\pi}{2}}\cos x\sin xdx = \int_0^1 udu = \left.\frac{u^2}{2}\right|_0^1 = \frac{1}{2}\]