# Reduction Formula for Integral of Power of Sine

## 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 $n$

$\blacksquare$