Calculus – Integration – The substitution rule

Theorem: If \(u=g(x)\) is a differentiable function whose range is an interval \(I\) and \(f\) is continuous on \(I\), then

\[\int f(g(x))g'(x)\,dx=\int f(u)\,du.\]

Note: If \(u=g(x)\), then: \(du=g'(x)\,dx\).

Theorem: If \(g'\) is continuous on \([a,b]\) and \(f\) is continuous on the range of \(u=g(x)\), then

\[\int_a^bf(g(x))g'(x)\,dx=\int_{g(a)}^{g(b)}f(u)\,du.\]

Definition: Let \(f\) be a function defined on a symmetric interval \(I\), then we have:

\(f\) is called even if \(f(-x)=f(x)\) for all \(x\in I\) \(\quad\)and\(\quad\) \(f\) is called odd if \(f(-x)=-f(x)\) for all \(x\in I\).

Theorem: Suppose that \(f\) is continuous on \([-a,a]\), then we have:

If \(f\) is even, then \(\displaystyle\int_{-a}^af(x)\,dx=2\int_0^af(x)\,dx\) \(\quad\)and\(\quad\) if \(f\) is odd, then \(\displaystyle\int_{-a}^af(x)\,dx=0\).

A nice formula with interesting applications

Using the substitution \(x=a+b-t\) or \(t=a+b-x\) we obtain that:

\[\int_a^b\frac{f(x)}{f(x)+f(a+b-x)}\,dx=-\int_b^a\frac{f(a+b-t)}{f(a+b-t)+f(t)}\,dt=\int_a^b\frac{f(a+b-t)}{f(a+b-t)+f(t)}\,dt.\]

This implies that:

\[I:=\int_a^b\frac{f(x)}{f(x)+f(a+b-x)}\,dx=\int_a^b\frac{f(a+b-x)}{f(a+b-x)+f(x)}\,dx.\]

Hence we have

\[2I=\int_a^b\frac{f(x)}{f(x)+f(a+b-x)}\,dx+\int_a^b\frac{f(a+b-x)}{f(a+b-x)+f(x)}\,dx=b-a,\]

which implies that

\[\int_a^b\frac{f(x)}{f(x)+f(a+b-x)}\,dx=\frac{b-a}{2}.\]

Some interesting applications are:

1. \(\displaystyle\int_0^1\frac{\sqrt{x}}{\sqrt{x}+\sqrt{1-x}}\,dx=\frac{1-0}{2}=\frac{1}{2},\quad\int_1^5\frac{\sqrt{x}}{\sqrt{x}+\sqrt{6-x}}\,dx=\frac{5-1}{2}=2,\quad\int_2^4\frac{\sqrt{x}}{\sqrt{x}+\sqrt{6-x}}\,dx=\frac{4-2}{2}=1\).


2. \(\displaystyle\int_0^{\frac{1}{2}\pi}\frac{\sin(x)}{\sin(x)+\cos(x)}\,dx=\int_0^{\frac{1}{2}\pi}\frac{\sin(x)}{\sin(x)+\sin(\frac{1}{2}\pi-x)}\,dx=\frac{\frac{1}{2}\pi-0}{2}=\frac{1}{4}\pi\).



Other similar examples are

3. \(\displaystyle\int_0^{\frac{1}{2}\pi}\frac{\sin^4(x)}{\sin^4(x)+\cos^4(x)}\,dx=\frac{1}{4}\pi,\quad\int_0^{\frac{1}{2}\pi}\frac{\sqrt{\sin(x)}}{\sqrt{\sin(x)}+\sqrt{\cos(x)}}\,dx=\frac{1}{4}\pi\).


4. \(\displaystyle\int_0^{\frac{1}{2}\pi}\frac{1}{1+\tan(x)}\,dx=\int_0^{\frac{1}{2}\pi}\frac{\frac{1}{\sin(x)}}{\frac{1}{\sin(x)}+\frac{1}{\cos(x)}}\,dx=\frac{1}{4}\pi\).

The Weierstrass substitution (Stewart §7.4, Exercise 59)

The Weierstrass substitution, named after the German mathematician Karl Weierstrass (1815-1897), converts an integral of a rational function in terms of \(\sin(x)\) and \(\cos(x)\) into an integral of an ordinary rational function.

For \(-\pi < x < \pi\) we set \(x=2\arctan(t)\) or \(t=\tan(\frac{1}{2}x)\). Then we have

\[\cos(\tfrac{1}{2}x)=\frac{1}{\sqrt{1+t^2}}\quad\text{and}\quad\sin(\tfrac{1}{2}x)=\frac{t}{\sqrt{1+t^2}},\]

which implies that

\[\sin(x)=2\sin(\tfrac{1}{2}x)\cos(\tfrac{1}{2}x)=\frac{2t}{1+t^2}\quad\text{and}\quad\cos(x)=2\cos^2(\tfrac{1}{2}x)-1=\frac{2}{1+t^2}-1=\frac{1-t^2}{1+t^2}.\]

Further we have

\[\frac{d}{dx}\tan(\tfrac{1}{2}x)=\frac{1}{2}\left(1+\tan^2(\tfrac{1}{2}x)\right)\quad\Longrightarrow\quad dx=\frac{2\,dt}{1+t^2}.\]

Some examples:

1) \[\int_0^{\frac{1}{2}\pi}\frac{dx}{1+\sin(x)}=\int_0^1\frac{1}{1+\frac{2t}{1+t^2}}\cdot\frac{2\,dt}{1+t^2}\,dt =\int_0^1\frac{2\,dt}{(1+t)^2}=-\frac{2}{1+t}\bigg|_0^1=-1+2=1.\]

2) \begin{align*} \int_0^{\frac{1}{2}\pi}\frac{\cos(x)}{1+\cos(x)}\,dx&=\int_0^1\frac{\frac{1-t^2}{1+t^2}}{1+\frac{1-t^2}{1+t^2}}\cdot\frac{2\,dt}{1+t^2} =\int_0^1\frac{1-t^2}{1+t^2+1-t^2}\cdot\frac{2\,dt}{1+t^2}=\int_0^1\frac{1-t^2}{1+t^2}\,dt\\[2.5mm] &=\int_0^1\frac{2-(1+t^2)}{1+t^2}\,dt=\bigg[2\arctan(t)-t\bigg]_0^1=\frac{1}{2}\pi-1. \end{align*}

3) \[\int_0^{\frac{1}{2}\pi}\frac{dx}{1+\sin(x)+\cos(x)}=\int_0^1\frac{1}{1+\frac{2t}{1+t^2}+\frac{1-t^2}{1+t^2}}\cdot\frac{2\,dt}{1+t^2} =\int_0^1\frac{dt}{1+t}=\bigg[\ln(1+t)\bigg]_0^1=\ln(2).\]

Serret's integral

Consider Serret's integral \(\displaystyle\int_0^1\frac{\ln(1+x)}{1+x^2}\,dx\). One way to evaluate the integral is by using the substitution \(x=\tan(\theta)\):

\[\int_0^1\frac{\ln(1+x)}{1+x^2}\,dx=\int_0^{\frac{1}{4}\pi}\ln\left(1+\tan(\theta)\right)\,d\theta=\int_0^{\frac{1}{4}\pi}\ln\left(\frac{\cos(\theta)+\sin(\theta)}{\cos(\theta)}\right)\,d\theta =\int_0^{\frac{1}{4}\pi}\left(\ln\left(\cos(\theta)+\sin(\theta)\right)-\ln\left(\cos(\theta)\right)\right)\,d\theta.\]

Now we use \(\cos(\theta)+\sin(\theta)=\sqrt{2}\cos(\theta-\frac{1}{4}\pi)\) to obtain

\[\int_0^1\frac{\ln(1+x)}{1+x^2}\,dx=\int_0^{\frac{1}{4}\pi}\left(\ln\left(\sqrt{2}\cos(\theta-\tfrac{1}{4}\pi)\right)-\ln\left(\cos(\theta)\right)\right)\,d\theta =\int_0^{\frac{1}{4}\pi}\left(\ln\left(\sqrt{2}\right)+\ln\left(\cos\left(\theta-\tfrac{1}{4}\pi\right)-\ln\left(\cos(\theta)\right)\right)\right)\,d\theta.\]

Finally, the substitution \(t=\frac{1}{4}\pi-\theta\) or \(\theta=\frac{1}{4}\pi-t\) shows that

\[\int_0^{\frac{1}{4}\pi}\ln\left(\cos\left(\theta-\tfrac{1}{4}\pi\right)\right)\,d\theta=-\int_{\frac{1}{4}\pi}^0\ln\left(\cos(t)\right)\,dt=\int_0^{\frac{1}{4}\pi}\ln\left(\cos(t)\right)\,dt,\]

which implies that

\[\int_0^1\frac{\ln(1+x)}{1+x^2}\,dx=\frac{1}{4}\pi\ln\left(\sqrt{2}\right)=\frac{1}{8}\pi\ln(2).\]

In this evaluation we obtained that

\[\int_0^{\frac{1}{4}\pi}\ln\left(\cos\left(\theta-\tfrac{1}{4}\pi\right)\right)\,d\theta=\int_0^{\frac{1}{4}\pi}\ln\left(\cos(t)\right)\,dt \quad\Longrightarrow\quad\int_0^{\frac{1}{4}\pi}\left(\ln\left(\cos\left(\theta-\tfrac{1}{4}\pi\right)-\ln\left(\cos(\theta)\right)\right)\right)\,d\theta=0.\]

The value of the integral \(\displaystyle\int_0^{\frac{1}{4}\pi}\ln\left(\cos(t)\right)\,dt\) is closely related to Catalan's constant, which will be considered later.

Another interesting example

Consider the integral \(\displaystyle\int_{\frac{1}{2}}^2\cos\left(x-\frac{1}{x}\right)\,dx\). Using the substitution \(x=\dfrac{1}{t}\) we obtain:

\[\int_{\frac{1}{2}}^2\cos\left(x-\frac{1}{x}\right)\,dx=\int_2^{\frac{1}{2}}\cos\left(\frac{1}{t}-t\right)\left(-\frac{1}{t^2}\right)\,dt =\int_{\frac{1}{2}}^2\frac{1}{t^2}\cos\left(t-\frac{1}{t}\right)\,dt.\]

This implies that

\[2\int_{\frac{1}{2}}^2\cos\left(x-\frac{1}{x}\right)\,dx=\int_{\frac{1}{2}}^2\left(1+\frac{1}{x^2}\right)\cos\left(x-\frac{1}{x}\right)\,dx.\]

Now we apply the substitution \(u=x-\dfrac{1}{x}\) and therefore \(du=\left(1+\frac{1}{x^2}\right)\,dx\):

\[\int_{\frac{1}{2}}^2\cos\left(x-\frac{1}{x}\right)\,dx=\frac{1}{2}\int_{-\frac{3}{2}}^{\frac{3}{2}}\cos(u)\,du =\frac{1}{2}\bigg[\sin(u)\bigg]_{-\frac{3}{2}}^{\frac{3}{2}}=\sin\left(\tfrac{3}{2}\right).\]

The Lambert \(W\) function

The Lambert \(W\) function is the inverse of the function \(y=xe^x\) with domain \([-1,\infty)\) and range \([-e^{-1},\infty)\). This implies that \(W(x)e^{W(x)}=x\).

Using the substitution \(t=W(x)\) or equivalently \(x=te^t\) and therefore \(dx=(t+1)e^t\,dt\) we have:

\[\int W(x)\,dx=\int t(t+1)e^t\,dt=(t^2-t+1)e^t+C=xW(x)-x+e^{W(x)}+C,\quad x>-e^{-1}.\]

For \(x>-e^{-1}\) and \(x\neq0\) this can also be written as: \(\displaystyle\int W(x)\,dx=x\left(W(x)-1+\frac{1}{W(x)}\right)+C\).

Since \(W(0)=0\) and \(W(e)=1\) it follows that \(\displaystyle\int_0^eW(x)\,dx=e-1\).

Similarly we obtain for \(x>-e^{-1}\) and \(x\neq0\):

\[\int\frac{W(x)}{x}\,dx=\int\frac{t}{te^t}(t+1)e^t\,dt=\int(t+1)\,dt=\frac{1}{2}t^2+t+C=\frac{1}{2}\{W(x)\}^2+W(x)+C.\]

Using \(W(x)e^{W(x)}=x\) this can also be obtained in this way:

\[\int\frac{W(x)}{x}\,dx=\int e^{-W(x)}\,dx=\int e^{-t}(t+1)e^t\,dt=\int(t+1)\,dt=\frac{1}{2}t^2+t+C=\frac{1}{2}\{W(x)\}^2+W(x)+C.\]

Since \(W(0)=0\) and \(W(e)=1\) it follows that \(\displaystyle\int_0^ee^{-W(x)}\,dx=\frac{3}{2}\).

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Last modified on March 10, 2024