Answer to Question #108786 in Calculus for Nimra

Question

∫tan²(ax) dx

Answer 1

Answer:$\int\tan^2ax\;dx\;=\;\frac1a\tan\;ax\;-\;x\;+\;C$

$\tan^2ax=\frac{\sin^2ax}{\cos^2ax}=\frac{1-\cos^2ax}{\cos^2ax}=sec^2ax-1\\\int\tan^2ax\;dx\;=\;\int(sec^2ax-1)\;dx\;=\\=\;\frac1a\int a\cdot sec^2ax\;dx\;-\;\int dx\\u\;=\;ax\;\Rightarrow\;du=a\;dx\\Substitute\;u.\\\int a\cdot sec^2ax\;dx\;=\;\int sec^2u\;du\;=\\=\;\tan u\;=\;\tan\;ax\\\frac1a\int a\cdot sec^2ax\;dx=\frac1a\tan\;ax\\\int dx=x\\\int\tan^2ax\;dx\;=\;\frac1a\tan\;ax\;-\;x\;+\;C$