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:

$\ds \int_0^{\frac \pi 2} \sin^2 x \rd x = \frac \pi 4$

$\blacksquare$


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