Primitive of Cube of Sine Function
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Theorem
- $\ds \int \sin^3 x \rd x = \frac {\cos^3 x} 3 - \cos x + C$
Proof 1
From Primitive of $\sin^3 a x$:
- $\ds \int \sin^3 a x \rd x = -\frac {\cos a x} a + \frac {\cos^3 a x} {3 a} + C$
The result follows by setting $a = 1$.
$\blacksquare$
Proof 2
\(\ds \int \sin^3 x \rd x\) | \(=\) | \(\ds \int \paren {\frac {3 \sin x - \sin 3 x} 4} \rd x\) | Power Reduction Formula for Cube of Sine | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 3 4 \int \sin x \rd x - \frac 1 4 \int \sin 3 x \rd x\) | Linear Combination of Primitives | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 3 4 \paren {-\cos x} - \frac 1 4 \paren {\frac {-\cos 3 x} 3 } + C\) | Primitive of $\sin x$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {-3 \cos x} 4 + \frac 1 {12} \paren {\cos 3 x} + C\) | simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {-3 \cos x} 4 + \frac 1 {12} \paren {4 \cos^3 x - 3 \cos x} + C\) | Triple Angle Formula for Cosine | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {-3 \cos x} 4 + \frac {\cos^3 x} 3 - \frac {\cos x} 4 + C\) | multipying out | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\cos^3 x} 3 - \cos x + C\) | simplifying |
$\blacksquare$
Sources
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text {II}$. Calculus: Exercises $\text {XIV}$: $9$.