Special Integral: \(\int_0^\frac{\pi}{2}\ln(\sin x)\;dx\)

I remember when I was in Year 12, we were given this question on an integration worksheet that challenged my friend, Parth, and I so much so that we stayed back after school to use the large whiteboard to solve this problem even after all our friends had left to go home. It is definitely one of my fondest memories of Mathematics at school and the feeling of triumph after finally completing it was absolutely rewarding.

This integral requires the special result:

\[\int_0^a f(x)\;dx = \int_0^a f(a-x)\;dx\]

It’ll also require trigonometric identities, logarithms, graphs and integration techniques.

Question

Find the value of:

\[\int_0^\frac{\pi}{2}\ln(\sin x)\;dx\]

Solution

Spoiler

Using the special result:

\[\begin{align*}I=&\int_0^\frac{\pi}{2}\ln(\sin x)\;dx\\I=&\int_0^\frac{\pi}{2}\ln\left(\sin \left(\frac{\pi}{2}-x\right)\right)\;dx\\I=&\int_0^\frac{\pi}{2}\ln(\cos x)\;dx\end{align*}\]

By adding the first and third lines above together:

\[\begin{align*}2I=&\int_0^\frac{\pi}{2}\left(\ln(\sin x)+\ln(\cos x)\right)\;dx\\=&\int_0^\frac{\pi}{2}\ln(\sin x \cos x)\;dx\\=&\int_0^\frac{\pi}{2}\ln\left(\frac{1}{2}\cdot 2\sin x\cos x\right)\;dx\\=&\int_0^\frac{\pi}{2}\ln\left(\frac{1}{2}\sin 2x\right)\;dx\end{align*}\]

We finally arrive at the following result (call it Equation *):

\[\begin{align}2I=&\int_0^\frac{\pi}{2}\left(\ln\left(\frac{1}{2}\right)+\ln(\sin 2x)\right)\;dx\end{align}\]

By looking at the graph of \(\ln(\sin 2x)\) in the domain \(0 < x < \frac{\pi}{2}\):

There is an axis of symmetry at \(x=\frac{\pi}{4}\), which means that:
\[\int_0^\frac{\pi}{2}\ln(\sin 2x)\;dx=2\int_0^\frac{\pi}{4}\ln(\sin 2x)\;dx\]

Using integration by substitution:

\[\begin{align*}u &= 2x\\du &= 2dx\end{align*}\]

When \(x=0,\;u=0\) and \(x=\frac{\pi}{4},\;u=\frac{\pi}{2}\).

\[2\int_0^\frac{\pi}{4}\ln(\sin 2x)\;dx=\int_0^\frac{\pi}{2}\ln(\sin u)\;du\]

Since the value of the definite integral doesn’t depend on what letter we use for the variable, i.e.:

\[\int_0^\frac{\pi}{2}\ln(\sin u)\;du=\int_0^\frac{\pi}{2}\ln(\sin x)\;dx\]

Hence:

\[\int_0^\frac{\pi}{2}\ln(\sin 2x)\;dx=\int_0^\frac{\pi}{2}\ln(\sin x)\;dx\]

Notice that the right hand side is actually the integral \(I\). Therefore, we can do the following in Equation *:

\[\begin{align*}2I &= \int_0^\frac{\pi}{2}\ln\left(\frac{1}{2}\right)\;dx + I\\ I & = \int_0^\frac{\pi}{2}\ln\left(\frac{1}{2}\right)\;dx\\&=\left. x\ln\left(\frac{1}{2}\right)\right|_0^\frac{\pi}{2}\\&=\frac{\pi}{2}\ln\left(\frac{1}{2}\right)\end{align*}\]

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