Reduction Formula for Integral of Power of Sine

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Theorem

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


Then:

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

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


Corollary

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


Proof 1

Let $n \ge 2$.

Let:

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


Then:

\(\ds I_n\) \(=\) \(\ds \int \sin^n x \rd x\)
\(\ds \) \(=\) \(\ds \int \sin^{n - 1} x \sin x \rd x\)
\(\ds \) \(=\) \(\ds \int \sin^{n - 1} x \map \rd {-\cos x}\) Derivative of Cosine Function
\(\ds \) \(=\) \(\ds - \sin^{n - 1} x \cos x - \int \paren {-\cos x} \map \rd {\sin^{n - 1} x}\) Integration by Parts
\(\ds \) \(=\) \(\ds - \sin^{n - 1} x \cos x - \int \paren {-\cos x} \paren {n - 1} \sin^{n - 2} x \cos x \rd x\) Derivative of Power and Chain Rule for Derivatives
\(\ds \) \(=\) \(\ds - \sin^{n - 1} x \cos x + \paren {n - 1} \int \sin^{n - 2} x \cos^2 x \rd x\) Linear Combination of Primitives and rearranging
\(\ds \) \(=\) \(\ds - \sin^{n - 1} x \cos x + \paren {n - 1} \int \sin^{n - 2} x \paren {1 - \sin^2 x} \rd x\) Sum of Squares of Sine and Cosine
\(\ds \) \(=\) \(\ds - \sin^{n - 1} x \cos x + \paren {n - 1} \int \sin^{n - 2} x \rd x - \paren {n - 1} \int \sin^n x \rd x\)
\(\ds \) \(=\) \(\ds - \sin^{n - 1} x \cos x + \paren {n - 1} I_{n - 2} - \paren {n - 1} I_n\)
\(\ds \leadsto \ \ \) \(\ds n I_n\) \(=\) \(\ds - \sin^{n - 1} x \cos x + \paren {n - 1} I_{n - 2}\) adding $\paren {n - 1} I_n$ to both sides
\(\ds \leadsto \ \ \) \(\ds I_n\) \(=\) \(\ds \dfrac {n - 1} n I_{n - 2} - \dfrac {\sin^{n - 1} x \cos x} n\) dividing by $n$ and rearranging

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


When $n = 1$ this degenerates to:

\(\ds \int \sin x \rd x\) \(=\) \(\ds \dfrac 0 1 \int \frac 1 {\sin x} \rd x - \dfrac {\sin^0 x \cos x} 1\)
\(\ds \) \(=\) \(\ds 0 - 1 \cdot \cos x\)
\(\ds \) \(=\) \(\ds -\cos x\)

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:

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

let:

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


and let:

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


Then:

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

$\blacksquare$


Also see


Sources