Definite Integral from 0 to Half Pi of Square of Sine x/Proof 2
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Theorem
- $\ds \int_0^{\frac \pi 2} \sin^2 x \rd x = \frac \pi 4$
Proof
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:
- $\ds \int_0^{\frac \pi 2} \sin^2 x \rd x = \frac \pi 4$
$\blacksquare$