 Solution

ID: past-exam-of-the-algebra-analysis-and-differential-equations-2-course-of-the-university-of-oxford/2004/5/ii/solution

For a nonnegative integer

n

, put

z = e^{i θ}

on the unit circle. Since

cos θ = (z + z^{- 1}) /2

and

d θ = d z / (i z)

\int_{0}^{2 π} cos^{2 n} θ d θ = \frac{1}{i 4 ^{n}} \int_{∣ z ∣ = 1} \frac{( z ^{2} + 1 ) ^{2 n}}{z ^{2 n + 1}} d z .

The residue at zero is the coefficient of

z^{2 n}

(1 + z^{2})^{2 n}

, namely

(n 2 n)

\int_{0}^{2 π} cos^{2 n} θ d θ = \frac{2 π}{4 ^{n}} (n 2 n) .

The contour integrand has no other poles. The formula also includes

n = 0

2 π

 New to topics? Read the docs here!