Definite Integral from 0 to Half Pi of Square of Sine x
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Theorem
- $\ds \int_0^{\frac \pi 2} \sin^2 x \rd x = \frac \pi 4$
Proof 1
\(\ds \int \sin^2 x \rd x\) | \(=\) | \(\ds \frac x 2 - \frac {\sin 2 x} 4 + C\) | Primitive of $\sin^2 x$ | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \int_0^{\frac \pi 2} \sin^2 x \rd x\) | \(=\) | \(\ds \intlimits {\frac x 2 - \frac {\sin 2 x} 4} 0 {\frac \pi 2}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\frac \pi 4 - \frac {\sin \pi} 4} - \paren {\frac 0 2 - \frac {\sin 0} 4}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac \pi 4\) | Sine of Multiple of Pi and simplifying |
$\blacksquare$
Proof 2
We have:
\(\ds \int_0^{\frac \pi 2} \sin^2 x \rd x\) | \(=\) | \(\ds \int_0^{\frac \pi 2} \sin^2 \paren {\frac \pi 2 - x} \rd x\) | Integral between Limits is Independent of Direction | |||||||||||
\(\ds \) | \(=\) | \(\ds \int_0^{\frac \pi 2} \cos^2 x \rd x\) | Sine of Complement equals Cosine |
So:
\(\ds 2 \int_0^{\frac \pi 2} \sin^2 x \rd x\) | \(=\) | \(\ds \int_0^{\frac \pi 2} \paren {\sin^2 x + \cos^2 x} \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int_0^{\frac \pi 2} \rd x\) | Sum of Squares of Sine and Cosine | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac \pi 2\) | Primitive of Constant |
giving:
- $\ds \int_0^{\frac \pi 2} \sin^2 x \rd x = \frac \pi 4$
$\blacksquare$
Also see
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 15$: Definite Integrals involving Trigonometric Functions: $15.29$