Reduction Formula for Definite Integral of Power of Sine

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Theorem

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

Let $I_n$ be defined as:

$\ds I_n = \int_0^{\frac \pi 2} \sin^n x \rd x$


Then $\sequence {I_n}$ is a decreasing sequence of real numbers which satisfies:

$n I_n = \paren {n - 1} I_{n - 2}$


Thus:

$I_n = \dfrac {n - 1} n I_{n - 2}$

is a reduction formula for $I_n$.


Corollary 1

\(\ds \int_0^{\frac \pi 2} \sin^{2 n} x \rd x\) \(=\) \(\ds \dfrac {\paren {2 n}!} {\paren {2^n n!}^2} \dfrac \pi 2\)
\(\ds \) \(=\) \(\ds \dfrac {1 \cdot 3 \cdot 5 \cdots \paren {2 n - 1} } {2 \cdot 4 \cdot 6 \cdots 2 n} \dfrac \pi 2\)

for $n \in \Z_{>0}$.


Corollary 2

\(\ds \int_0^{\frac \pi 2} \sin^{2 n + 1} x \rd x\) \(=\) \(\ds \dfrac {\paren {2^n n!}^2} {\paren {2 n + 1}!}\)
\(\ds \) \(=\) \(\ds \dfrac {2 \cdot 4 \cdot 6 \cdots 2 n} {3 \cdot 5 \cdot 7 \cdots \paren {2 n + 1} }\)

for $n \in \Z_{>0}$.


Proof

From Shape of Sine Function:

$\forall x \in \closedint 0 {\dfrac \pi 2}: 0 \le \sin x \le 1$

and so on the same interval:

$0 \le \sin^{n + 1} x \le \sin^n x$

therefore:

$\forall n \in \N: 0 < I_{n + 1} < I_n$


From Reduction Formula for Integral of Power of Sine:

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


Thus:

\(\ds I_n\) \(=\) \(\ds \int_0^{\frac \pi 2} \sin^n x \rd x\)
\(\ds \) \(=\) \(\ds \dfrac {n - 1} n \int_0^{\frac \pi 2} \sin^{n - 2} x \rd x - \intlimits {\dfrac {\sin^{n - 1} x \cos x} n} 0 {\pi / 2}\)
\(\ds \) \(=\) \(\ds \dfrac {n - 1} n I_{n - 2} - \frac {\paren {\sin^{n - 1} \frac \pi 2 \cos \frac \pi 2 - \sin^{n - 1} 0 \cos 0} } n\)
\(\ds \) \(=\) \(\ds \dfrac {n - 1} n I_{n - 2} - \frac {\paren {0 - \sin^{n - 1} 0 \cos 0} } n\) Cosine of Right Angle
\(\ds \) \(=\) \(\ds \dfrac {n - 1} n I_{n - 2} - \frac {\paren {0 - 0} } n\) Sine of Zero is Zero
\(\ds \leadsto \ \ \) \(\ds n I_n\) \(=\) \(\ds \paren {n - 1} I_{n - 2}\) multiplying both sides by $n$

$\blacksquare$


Also see


Sources

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