Why substitution works in indefinite integration

Let’s integrate \int{\frac{dx}{\sqrt{1-x^2}}} . We know the trick: substitute x for \sin\theta. We get dx=\cos\theta d\theta. Substituting into the original equation, we get \int{\frac{\cos\theta d\theta}{\sqrt{1-\sin^2\theta}}}=\int{\frac{\cos\theta d\theta}{|\cos\theta|}}. Let us assume \cos\theta remains positive throughout the interval under consideration. Then we get the integral as \theta or \arcsin x.

I have performed similar operations for close to five years of my life now. But I was never, ever, quite convinced with it. How can you, just like that, substitute dx for \cos\theta d\theta? My teacher once told me this: \frac{dx}{d\theta}=\cos\theta. Multiplying by d\theta on both sides, we get dx=d\theta. What?!! It doesn’t work like that!!

It was a year back that I finally derived why this ‘ruse’ works.

Take the function x^2. If you differentiate this with respect to x, you get 2x. If you integrate 2x, you get x^2+c. Simple.

Now take the function \sin^2\theta. Differentiate it with respect to \theta. You get 2\sin\theta.\cos\theta. If you integrate 2\sin\theta.\cos\theta, you get \sin^2\theta+c.

The thing to notice is when you integrate the two functions- 2x and 2\sin\theta.\cos\theta, you want a function of the form y^2. However and whatever I integrate, I ultimately want a function of the form y^2, so that I can substitute x for y to get x^2.

In the original situation, let us imagine there’s a function f(x)=\int{\frac{dx}{\sqrt{1-x^2}}}. We’ll discuss the properties of f(x). If we were to make the substitution x=\sin\theta in f(x) and differentiate it with respect to \theta, we’d get a function of the form \frac{1}{\sqrt{1-y^2}}\cos\theta, where y is \sin\theta. There are two things to note here:

1. The form of the derivative if f(x) wrt \theta is the same as that of f'(x), which is \frac{1}{\sqrt{1-y^2}}, multiplied by \cos\theta, or derivative of \sin\theta wrt \theta.

2. When any function is differentiated with respect to any variable, integration wrt the same variabe gives us back the same function. Hence, \int{\frac{\partial f}{\partial x}dx}=\int{\frac{\partial f}{\partial \theta}d\theta}

Coming back to \int{\frac{dx}{\sqrt{1-x^2}}}, let us assume its integral is f(x). It’s derivative on substituting x=\cos\theta and differentiating wrt \theta is of the same form as \frac{\partial f}{\partial x} multiplied by \cos\theta. This is a result of the chain rule of differentiation. Now following rule 2, we know \int{\frac{dx}{\sqrt{1-x^2}}}=\int{\frac{\cos\theta d\theta}{\sqrt{1-\sin^2\theta}}}.

How is making the substitution x=\sin\theta justified? Could we have made any other continuous substitution, like x=\theta^2 +\tan\theta^3? Let us assume we substitute x for g(\theta). We want g(\theta) to take all the values x can take. This is the condition that must be satisfied by any substitution. For values that g(\theta) takes by x doesn’t, we restrict the range of g(\theta) to that of x. Note that the shapes of f(x) as plotted against x and f(\sin\theta) as plotted against \theta will be different. But that is irrelevant as long as we can write the same cartesian pairs (m,n) for any variable, where m is the x-coordinate and n is the y-coordinate.

Summing the argument, we predict the form the derivative of f(x) will take when the substitution x=\sin\theta is made, and then integrate this new form wrt \theta to get the original function. This is why the ‘trick’ works.

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Graduate student

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