Reduction Formula for Definite Integral of Power of Cosine

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $n \in \Z_{> 0}$ be a positive integer.

Let $I_n$ be defined as:

$\displaystyle I_n = \int_0^{\frac \pi 2} \cos^n x \rd x$


Then $\left\langle{I_n}\right\rangle$ is a decreasing sequence of real numbers which satisfies:

$n I_n = \left({n - 1}\right) I_{n - 2}$


Thus:

$I_n = \dfrac {n - 1} n I_{n - 2}$

is a reduction formula for $I_n$.


Corollary 1

$\displaystyle \int_0^{\frac \pi 2} \cos^{2 n} x \rd x = \dfrac {\left({2 n}\right)!} {\left({2^n n!}\right)^2} \dfrac \pi 2$


Corollary 2

$\displaystyle \int_0^{\frac \pi 2} \cos^{2 n + 1} x \rd x = \dfrac {\left({2^n n!}\right)^2} {\left({2 n + 1}\right)!}$


Proof

From Shape of Cosine Function:

$\forall x \in \left[{0 \,.\,.\, \dfrac \pi 2}\right]: 0 \le \cos x \le 1$

and so on the same interval:

$0 \le \cos^{n + 1} x \le \cos^n x$

therefore:

$\forall n \in \N: 0 < I_{n + 1} < I_n$


From Reduction Formula for Integral of Power of Cosine:

$\displaystyle \int \cos^n x \rd x = \dfrac {\cos^{n - 1} x \sin x} n + \dfrac {n - 1} n \int \cos^{n - 2} x \rd x$


Thus:

\(\displaystyle I_n\) \(=\) \(\displaystyle \int_0^{\frac \pi 2} \cos^n x \rd x\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {n - 1} n \int_0^{\frac \pi 2} \cos^{n - 2} x \rd x - \left[{\dfrac {\cos^{n - 1} x \sin x} n}\right]_0^{\pi / 2}\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {n - 1} n I_{n - 2} - \frac {\left({\cos^{n - 1} \frac \pi 2 \sin \frac \pi 2 - \cos^{n - 1} 0 \sin 0}\right)} n\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {n - 1} n I_{n - 2} - \frac {\left({0 - \cos^{n - 1} 0 \sin 0}\right)} n\) Cosine of Right Angle
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {n - 1} n I_{n - 2} - \frac {\left({0 - 0}\right)} n\) Sine of Zero is Zero
\(\displaystyle \implies \ \ \) \(\displaystyle n I_n\) \(=\) \(\displaystyle \left({n - 1}\right) I_{n - 2}\) multiplying both sides by $n$

$\blacksquare$


Also see