Primitive of Square of Sine of a x

Theorem

$\ds \int \sin^2 a x \rd x = \frac x 2 - \frac {\sin 2 a x} {4 a} + C$

$\ds \int \sin^2 x \rd x$

$=$

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

$\ds \leadsto \ \ $

$\ds \int \sin^2 a x \rd x$

$=$

$\ds \frac 1 a \paren {\frac {a x} 2 - \frac {\sin 2 a x} 4} + C$

$\ds $

$=$

$\ds \frac x 2 - \frac {\sin 2 a x} {4 a} + C$

simplifying

$\blacksquare$

1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $\sin a x$: $14.347$
1968: George B. Thomas, Jr.: Calculus and Analytic Geometry (4th ed.) ... (previous) ... (next): Front endpapers: A Brief Table of Integrals: $58$.