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


Author: ficetea

I'll write more here later!

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