Reduction Formula for Integral of Power of Sine/Proof 2

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

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 $n$

$\blacksquare$