Answer to Question #108782 in Calculus for Nimra

Question

∫1/cos(ax) dx

Answer 1

Answer:$\int\frac1{\cos\;ax}dx=\frac1a\ln\left|sec\;ax\;+\;\tan\;ax\right|+C$

$\int\frac1{\cos\;ax}dx=\;\frac1a\int a\;\frac1{\cos\;ax}\;dx\\u=ax\;\Rightarrow\;du\;=\;a\;dx\\Substitute\;u.\\\frac1a\int a\;\frac1{\cos\;ax}\;dx=\frac1a\int\frac1{\cos u}\;du=\\=\frac1a\int sec\;u\;du\\\int sec\;u\;du=\int sec\;u\;\frac{sec\;u\;+\;\tan\;u}{sec\;u\;+\;\tan\;u}\;du\;=\\=\int\frac{(sec^2u+sec\;u\;\tan\;u)\;du}{sec\;u\;+\;\tan\;u}\\v=sec\;u\;+\;\tan\;u\\dv\;=\;(sec\;u\;\tan\;u\;+\;sec^2u)\;du\\Substitute\;v.\\\int\frac{(sec^2u+sec\;u\;\tan\;u)\;du}{sec\;u\;+\;\tan\;u}=\int\frac{dv}v=\\=\ln\left|v\right|+C=\ln\left|sec\;u\;+\;\tan\;u\right|+C\;=\\=\;\ln\left|sec\;ax\;+\;\tan\;ax\right|+C\\\; \int\frac1{\cos\;ax}dx=\frac1a\ln\left|sec\;ax\;+\;\tan\;ax\right|+C$