Trigonometric substitution in integration

Why trigonometric substitution in integration: this is something that puzzled me and made me hate differentiation/integration during my IITJEE preparation. A lot of the techniques that we learned were based on algorithmic memorization rather than a feel for what really was happening. Thankfully, I have come to a college that requires 0% attendance, so that I can fulfill all my deepest desires for understanding.

Say we have \int_{0}^{1}{\frac{1}{\sqrt{1-x^2}}dx}. A very common method of integrating this function is to make the substitution x=\sin\theta. Moreover, dx=\cos\theta d\theta. Hence, we get \int_0^{\frac{\pi}{2}}{\frac{\cos\theta }{|\sin\theta|}d\theta}.

Today we have to understand why that works.

When we say x=\sin\theta, what we’re saying is for some variable \theta, every number in the interval [0,1] is the image of some value of \theta. In others words, there is a bijective mapping \sin\theta:\theta\to[0,1]. Hence if we replace some value of x with the sine of the corresponding \theta, we should not find a difference.

However, there is the matter of dx. The function \int{f(x)dx} is nothing but the limit of making the intervals \Delta x on the x-axis smaller and smaller, and finding the summation \sum{f(x)\Delta x} (of course this limit is valid only if the same limit is reached regardless of the x we take in the interval). Now as x=\sin\theta, we have \Delta x=|x_i-x_j|=\cos\theta\Delta\theta for some \theta\in[\arcsin{x_i},\arcsin{x_j}], by the Mean Value Theorem. Moreover, the limits of integration will also change, but this is obvious.

We ultimately get \int_0^{\frac{\pi}{2}}{\frac{\cos\theta }{|\sin\theta|}d\theta}.

What we have in essence is a sequence of infinite summations, approaching a limit (the integral). Say one of the terms of the sequence is \frac{1}{2}*\frac{1}{2}+\frac{1}{4}*\frac{1}{4}+\dots. We can write this very summation in terms of another function (substitution x for \cos\theta. That is all that we’re doing here.

I hope this article helps those who ask “why”, and suffer for it.

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Graduate student

2 thoughts on “Trigonometric substitution in integration

  1. Even I study in an institute that requires 0% attendance but never realized that I could use this to explore and increase my level of understanding

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