Reduction Formula for Integral of Power of Sine/Proof 1

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 $\ds \int \sin^n x \rd x$.

Proof

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 $n = 1$.

$\blacksquare$

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