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Tangent half-angle substitution is a technique primarily utilised in integral calculus.

## The substitution

### Differential

First, let $t = \tan\left(\frac{x}{2}\right)$ Differentiate using the chain rule

$\frac{\mathrm{d}t}{\mathrm{d}x} \underbrace{=}_{\text{let } u = x/2} \frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{x}{2}\right) \frac{\mathrm{d}}{\mathrm{d}u} \tan(u) = \frac{1}{2}\sec^2(u) = \frac{1}{2}\sec^2\left(\frac{x}{2}\right)$.

Consider the Pythagorean identity

$1 + \tan^2(x) = \sec^2(x)$

Therefore,

$\frac{\mathrm{d}t}{\mathrm{d}x} = \frac{1}{2}\sec^2\left(\frac{x}{2}\right) = \frac{1+t^2}{2}$

Isolate $\mathrm{d}x$,

$\frac{\mathrm{d}x}{\mathrm{d}t} = \frac{2}{1+t^2}$
$\mathrm{d}x = \frac{2}{1+t^2}\mathrm{d}t$

### Sine

Consider the double angle identity,

$\sin(x) = 2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)$

Which may be written as,

$\sin(x) = 2\frac{\sin(\frac{x}{2})}{\cos(\frac{x}{2})}\cos^2\left(\frac{x}{2}\right)$

Which in turn may be written as,

$\sin(x) = 2t\cos^2\left(\frac{x}{2}\right)$

Recall that,

$\sec(x) = \frac{1}{\cos(x)}$

We can therefore rewrite the above as,

$\sin(x) = \frac{2t}{\sec^2(\frac{x}{2})}$

Therefore,

$\sin(x) = \frac{2t}{1+t^2}$.

### Cosine

Consider the double angle identity,

$\cos(x) = 2\cos^2\left(\frac{x}{2}\right)-1$.

Rewriting,

$\cos(x) = \frac{2}{\sec^2(\frac{x}{2})} - 1$
$\cos(x) = \frac{2-\sec^2(\frac{x}{2})}{\sec^2(\frac{x}{2})}$

Writing in terms of t,

$\cos(x) = \frac{2-(1+t^2)}{1+t^2} = \frac{1-t^2}{1+t^2}$.

## Examples

### Example 1

$\mathrm{I}_1 = \int \frac{1}{2+\cos(x)} \mathrm{d}x$

Let,

$t = \tan\left(\frac{x}{2}\right)$
$\implies \mathrm{d}x = \frac{2}{1+t^2}$.

Using,

$\cos(x) = \frac{1-t^2}{1+t^2}$.

We may rewrite the integral as,

$\int \frac{1}{2+\frac{1-t^2}{1+t^2}} \cdot \frac{2}{1+t^2} \mathrm{d}t$.
$\int \frac{1}{2+2t^2 + 1 - t^2} \mathrm{d}t$.

From there,

$\mathrm{I}_1 = \int \frac{1}{t^2+3} \mathrm{d}t$.

Use the substitution,

$t = \sqrt{3}\tan(\alpha)$
$\mathrm{d}t = \sqrt{3}\sec^2(\alpha)\mathrm{d}\alpha$.

Therefore,

$\mathrm{I}_1 = \int \frac{\sqrt{3}\sec^2(\alpha)\mathrm{d}\alpha}{3\tan^2(\alpha)+3}$
$\mathrm{I}_1 = \frac{\sqrt{3}}{3} \int \frac{\sec^2(\alpha)}{\sec^2(\alpha)}\mathrm{d}\alpha$
$\mathrm{I}_1 = \frac{1}{\sqrt{3}} \int \mathrm{d}\alpha$
$\mathrm{I}_1 = \frac{1}{\sqrt{3}} \alpha + C$

Isolate $\alpha$,

$\frac{t}{\sqrt{3}} = \tan(\alpha)$
$\arctan\left(\frac{t}{\sqrt{3}}\right) = \alpha$

So,

$\mathrm{I}_1 = \frac{1}{\sqrt{3}} \arctan\left(\frac{t}{\sqrt{3}}\right) + C$
$\mathrm{I}_1 = \frac{1}{\sqrt{3}} \arctan\left(\frac{\tan(\frac{x}{2})}{\sqrt{3}}\right) + C$

### Example 2

$\sin(x) + \cos(x) = 1$

Let $t = \tan\left(\frac{x}{2}\right)$

$\frac{2t}{1+t^2} + \frac{1-t^2}{1+t^2} = 1$

Multiplying through $1+t^2$

$2t + 1 - t^2 = 1+t^2$

Rearranging,

$2t - 2t^2 = 0$
$\implies t = 0 \text{ or } t = 1$.

First case,

$\tan\left(\frac{x}{2}\right) = 0$

Apply arctangent to both sides

$\frac{x}{2} = \arctan(0) + n\pi = 0 + n\pi = n\pi$
$x = 2n\pi$

Second case,

$\tan\left(\frac{x}{2}\right) = 1$

Again, apply arctangent to both sides,

$\frac{x}{2} = \arctan(1) + n\pi = \frac{\pi}{4} + n\pi = \frac{\pi + 4n\pi}{4} = \frac{(4n+1)\pi}{4}$
$x = \frac{(4n+1)\pi}{2}$

So the two general solutions are,

$x = 2n\pi$
$x = \frac{(4n+1)\pi}{2}$

Where $n$ is any integer.