Answer to Question #108783 in Calculus for Nimra

Question

∫sin²(ax)dx

Answer 1

Answer:$\int\sin^2(ax)\;dx\;=\frac12x-\frac{\sin\;2ax}{4a}+C$

$\int\sin^2(ax)\;dx=\;\frac1a\int a\;\sin^2(ax)\;dx\\u=ax\;\Rightarrow\;du\;=\;a\;dx\\Substitute\;u.\\\frac1a\int a\;\sin^2(ax)\;dx=\frac1a\int\sin^2u\;du\\\sin^2u=\;\frac{1-\cos2u}2\\\int\sin^2u\;du\;=\;\int(\frac12-\frac{\cos2u}2)\;du\;=\;\\=\frac12u-\frac12\times\frac{\sin2u}2+C=\frac12u-\frac{\sin2u}4+C=\\=\frac12ax-\frac{\sin2ax}4+C\\\int\sin^2(ax)\;dx\;=\frac1a(\;\frac12ax-\frac{\sin2ax}4)+C=\\=\frac12x-\frac{\sin2ax}{4a}+C$