Reduction Formula for Integral of Power of Cosine

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Theorem

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

Let:

$I_n := \ds \int \cos^n x \rd x$

Then:

$I_n = \dfrac {\cos^{n - 1} x \sin x} n + \dfrac {n - 1} n I_{n - 2}$

is a reduction formula for $\ds \int \cos^n x \rd x$.


Corollary

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


Proof

With a view to expressing the problem in the form:

$\ds \int u \frac {\rd v} {\rd x} \rd x = u v - \int v \frac {\rd u} {\rd x} \rd x$

let:

\(\ds u\) \(=\) \(\ds \cos^{n - 1} x\)
\(\ds \leadsto \ \ \) \(\ds \frac {\rd u} {\rd x}\) \(=\) \(\ds -\paren {n - 1} \cos ^{n - 2} x \sin x\) Chain Rule for Derivatives, Derivative of Cosine Function, Derivative of Power


and let:

\(\ds \frac {\rd v} {\rd x}\) \(=\) \(\ds \cos x\)
\(\ds \leadsto \ \ \) \(\ds v\) \(=\) \(\ds \sin x\) Primitive of Cosine Function


Then:

\(\ds I_n\) \(=\) \(\ds \int \cos^n x \rd x\)
\(\ds \) \(=\) \(\ds \int \cos^{n - 1} x \cos x \rd x\)
\(\ds \) \(=\) \(\ds \cos^{n - 1} x \sin x - \int \sin x \paren {-\paren {n - 1} \cos^{n - 2} x \sin x} \rd x\) Integration by Parts
\(\ds \) \(=\) \(\ds \cos^{n - 1} x \sin x + \int \paren {n - 1} \cos^{n - 2} x \sin^2 x \rd x\) rearranging
\(\ds \) \(=\) \(\ds \cos^{n - 1} x \sin x + \int \paren {n - 1} \cos^{n - 2} x \paren {1 - \cos^2 x} \rd x\) Sum of Squares of Sine and Cosine
\(\ds \) \(=\) \(\ds \cos^{n - 1} x \sin x + \paren {n - 1} \int \cos^{n - 2} x \rd x - \int \paren {n - 1} \cos^n x \rd x\) Linear Combination of Primitives
\(\ds \) \(=\) \(\ds \cos^{n - 1} x \sin x + \paren {n - 1} I_{n - 2} - \paren {n - 1} I_n\) substituting for $I_n$ and $I_{n - 2}$
\(\ds \leadsto \ \ \) \(\ds n I_n\) \(=\) \(\ds \cos^{n - 1} x \sin x + \paren {n - 1} I_{n - 2}\) rearranging
\(\ds \leadsto \ \ \) \(\ds I_n\) \(=\) \(\ds \frac {\cos^{n - 1} x \sin x} n + \dfrac {n - 1} n I_{n - 2}\) dividing both sides by $n$

$\blacksquare$


Also see


Sources