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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.

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