Half Angle Substitution

Here’s a neat substitution for integration that I never learned from calculus subjects but found in a book. I also frequently forget the formulae so whenever I want to use it I need to re-derive everything, hopefully writing it here will help me remember.

If x is our integration variable, then we make the substitution x = \tan(t/2). Using the trig identities, we can now get expressions for trigonometric functions in terms of t.

\sin x = 2\sin \frac{x}{2}\cos \frac{x}{2}
= 2\tan\frac{x}{2}\cos^2\frac{x}{2}
= 2\frac{t}{\sec^2(x/2)}
= 2t/(t^2+1).

\cos x = 1 - 2\sin^2\frac{x}{2}
= 1 - 2\tan^2\frac{x}{2}\cos^2\frac{x}{2}
= (t^2+1-2t^2)/(t^2+1)
= (1-t^2)/(1+t^2)

\tan x = \sin x /\cos x
= 2t/(1-t^2)

And of course dt = \frac{1}{2}\sec^2\frac{x}{2}dx so dx = 2dt/(1+t^2).
Now we can work out things like \int \csc x dx, which is given by
\int dt/t = \ln\tan\frac{x}{2} + c


Happy Belated Pi Day

I didn’t create this blog until after Pi day, so unfortunately I’m a bit late. I celebrate Pi day by deriving the same series expression for pi every year. I’m not sure who to attribute this particular method to, but I definitely learned in from some old calculus book once upon a time. It’s as simple as calculating \int_0^1\frac{1}{x^2+1}dx two different ways and equating the results. First up, we can make the substitution x = \tan\theta and simplify yielding \int_0^{\pi/4}d\theta = \frac{\pi}{4}. On the other hand, if |x|<1 then \sum_{n=0}^\infty(-1)^nx^{2n} converges (uniformly on all intervals |x|\leq r < 1) to \frac{1}{x^2+1}, and so may proceed with

\int_0^1\frac{1}{1+x^2}dx = \sum_{n=0}^\infty(-1)^n\int_0^1x^{2n}dx. Each \int_0^1x^{2n}dx = \frac{1}{2n+1}, and so putting it all together we’re left with

\pi = 4\sum_{n=0}^\infty \frac{(-1)^n}{2n+1} = 4(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+...)

And there it is!