Definite Integral from 0 to Half Pi of Square of Sine x/Proof 1

Theorem

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

Proof

\(\ds \int \sin^2 x \rd x\)

\(=\)

\(\ds \frac x 2 - \frac {\sin 2 x} 4 + C\)

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

$\blacksquare$