Definite Integral from 0 to Half Pi of Odd Power of Cosine x/Proof 1
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Theorem
- $\ds \int_0^{\frac \pi 2} \cos^{2 n + 1} x \rd x = \dfrac {\paren {2^n n!}^2} {\paren {2 n + 1}!}$
Proof
The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
- $\ds \int_0^{\frac \pi 2} \cos^{2 n + 1} x \rd x = \dfrac {\paren {2^n n!}^2} {\paren {2 n + 1}!}$
Basis for the Induction
$\map P 0$ is the case:
\(\ds \int_0^{\frac \pi 2} \cos x \rd x\) | \(=\) | \(\ds \bigintlimits {\sin x} 0 {\frac \pi 2}\) | Primitive of $\cos x$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \sin \frac \pi 2 - \sin 0\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1\) | Sine of Right Angle, Sine of Zero is Zero | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {2^0 \times 1!}^2} {\paren {2 \times 0 + 1}!}\) |
Thus $\map P 0$ is seen to hold.
This is the basis for the induction.
Induction Hypothesis
Now it needs to be shown that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
- $\ds \int_0^{\frac \pi 2} \cos^{2 k + 1} x \rd x = \dfrac {\paren {2^k k!}^2} {\paren {2 k + 1}!}$
from which it is to be shown that:
- $\ds \int_0^{\frac \pi 2} \cos^{2 \paren {k + 1} + 1} x \rd x = \dfrac {\paren {2^{k + 1} \paren {k + 1}!}^2} {\paren {2 \paren {k + 1} + 1}!}$
Induction Step
This is the induction step:
Let $I_k = \ds \int_0^{\frac \pi 2} \cos^{2 k + 1} x \rd x$.
\(\ds I_{k + 1}\) | \(=\) | \(\ds \frac {2 \paren {k + 1} } {2 \paren {k + 1} + 1} I_k\) | Reduction Formula for Definite Integral of Power of Cosine | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {2 \paren {k + 1} } {2 \paren {k + 1} + 1} \dfrac {\paren {2^k k!}^2} {\paren {2 k + 1}!}\) | Induction Hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {2 \paren {k + 1} }^2} {\paren {2 \paren {k + 1} + 1} \paren {2 \paren {k + 1} } } \dfrac {\paren {2^k k!}^2} {\paren {2 k + 1}!}\) | multiplying top and bottom by $2 \paren {k + 1}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\paren {2^{k + 1} \paren {k + 1}!}^2} {\paren {2 \paren {k + 1} + 1}!}\) |
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
- $\ds \forall n \in \Z_{\ge 0}: \int_0^{\frac \pi 2} \cos^{2 n + 1} x \rd x = \dfrac {\paren {2^n n!}^2} {\paren {2 n + 1}!}$
$\blacksquare$