Answer to Question #108784 in Calculus for Nimra

Question

∫cos²(ax)dx

Answer 1

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

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