Reduction Formula for Integral of Power of Sine

From ProofWiki
Jump to navigation Jump to search

Theorem

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


Then:

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

is a reduction formula for $\displaystyle \int \sin^n x \rd x$.


Corollary

$\displaystyle \int \sin^n a x \ \mathrm d x = \dfrac {n - 1} n \int \sin^{n - 2} a x \ \mathrm d x - \dfrac {\sin^{n-1} a x \cos a x} {a n}$


Proof 1

Let $n \ge 2$.

Let:

$\displaystyle I_n := \int \sin^n x \ \mathrm d x$


Then:

\(\displaystyle I_n\) \(=\) \(\displaystyle \int \sin^n x \ \mathrm d x\)
\(\displaystyle \) \(=\) \(\displaystyle \int \sin^{n-1} x \sin x \ \mathrm d x\)
\(\displaystyle \) \(=\) \(\displaystyle \int \sin^{n-1} x \ \mathrm d \left({-\cos x}\right)\) Derivative of Cosine Function
\(\displaystyle \) \(=\) \(\displaystyle - \sin^{n-1} x \cos x - \int \left({-\cos x}\right) \mathrm d \left({\sin^{n-1} x }\right)\) Integration by Parts
\(\displaystyle \) \(=\) \(\displaystyle - \sin^{n-1} x \cos x - \int \left({-\cos x}\right) \left({n - 1}\right)\sin^{n-2} x \cos x \ \mathrm d x\) Derivative of Power and Chain Rule
\(\displaystyle \) \(=\) \(\displaystyle - \sin^{n-1} x \cos x + \left({n - 1}\right) \int \sin^{n-2} x \cos^2 x \ \mathrm d x\) Linear Combination of Integrals and rearranging
\(\displaystyle \) \(=\) \(\displaystyle - \sin^{n-1} x \cos x + \left({n - 1}\right) \int \sin^{n-2} x \left({1 - \sin^2 x}\right) \ \mathrm d x\) Sum of Squares of Sine and Cosine
\(\displaystyle \) \(=\) \(\displaystyle - \sin^{n-1} x \cos x + \left({n - 1}\right) \int \sin^{n-2} x \ \mathrm d x - \left({n - 1}\right) \int \sin^n x \ \mathrm d x\)
\(\displaystyle \) \(=\) \(\displaystyle - \sin^{n-1} x \cos x + \left({n - 1}\right) I_{n-2} - \left({n - 1}\right) I_n\)
\(\displaystyle \implies \ \ \) \(\displaystyle n I_n\) \(=\) \(\displaystyle - \sin^{n-1} x \cos x + \left({n - 1}\right) I_{n-2}\) adding $\left({n - 1}\right) I_n$ to both sides
\(\displaystyle \implies \ \ \) \(\displaystyle I_n\) \(=\) \(\displaystyle \dfrac {n - 1} n I_{n-2} - \dfrac {\sin^{n-1} x \cos x} n\) dividing by $n$ and rearranging

thus demonstrating the identity for all $n \ge 2$.


When $n = 1$ this degenerates to:

\(\displaystyle \int \sin x \ \mathrm d x\) \(=\) \(\displaystyle \dfrac 0 1 \int \frac 1 {\sin x} \ \mathrm d x - \dfrac {\sin^0 x \cos x} 1\)
\(\displaystyle \) \(=\) \(\displaystyle 0 - 1 \cdot \cos x\)
\(\displaystyle \) \(=\) \(\displaystyle -\cos x\)

From Primitive of Sine Function this shows that the identity still holds for $n = 1$.

$\blacksquare$


Proof 2

With a view to expressing the problem in the form:

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

let:

\(\displaystyle u\) \(=\) \(\displaystyle \sin^{n - 1} x\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \frac {\d u} {\d x}\) \(=\) \(\displaystyle \paren {n - 1} \sin ^{n - 2} x \cos x\) Chain Rule, Derivative of Sine Function, Derivative of Power


and let:

\(\displaystyle \frac {\d v} {\d x}\) \(=\) \(\displaystyle \sin x\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle v\) \(=\) \(\displaystyle -\cos x\) Primitive of Sine Function


Then:

\(\displaystyle \int \sin^n x \rd x\) \(=\) \(\displaystyle \int \sin^{n - 1} x \sin x \rd x\)
\(\displaystyle \) \(=\) \(\displaystyle \sin^{n - 1} x \paren {-\cos x} - \int \paren {-\cos x} \paren {\paren {n - 1} \sin^{n - 2} x \cos x} \rd x\) Integration by Parts
\(\displaystyle \) \(=\) \(\displaystyle \int \paren {n - 1} \sin^{n - 2} x \cos^2 x \rd x - \sin^{n - 1} x \cos x\) rearranging
\(\displaystyle \) \(=\) \(\displaystyle \int \paren {n - 1} \sin^{n - 2} x \paren {1 - \sin^2 x} \rd x - \sin^{n - 1} x \cos x\) Sum of Squares of Sine and Cosine
\(\displaystyle \) \(=\) \(\displaystyle \paren {n - 1} \int \sin^{n - 2} x \rd x - \paren {n - 1} \int \sin^n x \rd x - \sin^{n - 1} x \cos x\) Linear Combination of Integrals
\(\displaystyle \leadsto \ \ \) \(\displaystyle n \int \sin^n x \rd x\) \(=\) \(\displaystyle \paren {n - 1} \int \sin^{n - 2} x \rd x - \sin^{n - 1} x \cos x\) rearranging
\(\displaystyle \leadsto \ \ \) \(\displaystyle \int \sin^n x \rd x\) \(=\) \(\displaystyle \dfrac {n - 1} n \int \sin^{n - 2} x \rd x - \dfrac {\sin^{n - 1} x \cos x} n\) dividing both sides by $n$

$\blacksquare$


Also see


Sources