Tangent half angle subsititution (Integration Query)

Question

How would you go about solving something like $$\int \frac{1}{4+5\cos{t}}~dt$$

I am aware that you would use a substitution involving the half angle of tan but I am unsure on how to apply it.

Many thanks

Answer 1

This is also known as Weierstrass Substitution.

This is done by substituting $u=\tan\left(\frac{t}{2}\right)$. We must derive formulae in terms of $u$ for $\cos{t}$ and $dt$. The substitutions are rather standard, but I will derive them for you:

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For $\cos{t}$, we use the identity: $$\cos{\theta}\equiv 1-2\sin^2\left(\frac{\theta}{2}\right) \tag{1}$$ Applying it gives: $$\cos{t}= 1-2\sin^2\left(\frac{t}{2}\right)=1-\frac{2\tan^2\left(\frac{t}{2}\right)}{\sec^2\left(\frac{t}{2}\right)}$$ Now, we use the following identity on $\sec^2\left(\frac{t}{2}\right)$: $$\tan^2{\theta}+1\equiv\sec^2{\theta} \tag{2}$$ This gives: $$\cos{t}=1-\frac{2u^2}{u^2+1}$$ Hence: $$\boxed{\cos{t}=\frac{1-u^2}{u^2+1}}$$

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Now, for $dt$, we first evaluate the derivative of $u$ with respect to $t$: $$\frac{du}{dt}=\frac{1}{2}\sec^2\left(\frac{t}{2}\right)$$ Using the identity in $(2)$, we obtain: $$\frac{du}{dt}=\frac{1}{2}(u^2+1)$$ Hence: $$\boxed{dt=\frac{2}{u^2+1}~du}$$

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After substitution, we obtain: $$\int \frac{1}{4+5\cos{t}}~dt=\int \frac{1}{4+5\left(\frac{1-u^2}{u^2+1}\right)}\cdot \frac{2}{u^2+1}~du=\int \frac{2}{9-u^2}~du$$ Can you continue?

Answer 2

use the substitution $$t=\tan\left(\frac{x}{2}\right)$$ and $$\cos(x)={\frac {1- \left( \tan \left( t/2 \right) \right) ^{2}}{1+ \left( \tan \left( t/2 \right) \right) ^{2}}} $$