Using Substitution To Establish an Integration Formula

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1. \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ \int \tan x\,dx=-\ln \left\vert \cos x\right\vert +C=\ln \vert \sec x\vert +C }} \]
2. \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ \int \sec x\,dx=\ln \vert \sec x+\tan x\vert +C }} \]

Solution (a) Since \(\tan x=\dfrac{\sin x}{\cos x},\) we let \(u=\cos x.\) Then \(du=-\sin x\, dx\) and \[ \begin{eqnarray*} \int \tan x dx = \int \dfrac{\sin x}{\cos x}dx=\int -\dfrac{du}{u}=-\ln \left\vert u\right\vert +C=-\ln \left\vert \cos x\right\vert +C \\ \underset{ \underset{ \color{#0066A7}{\hbox{\(r\ln x=\ln x^{r}\)}} } {\color{#0066A7}{\uparrow}} } {=} \ln \left\vert \cos x\right\vert^{-1}+C =\ln \left\vert \dfrac{1}{\cos x}\right\vert + C =\ln \left\vert \sec x\right\vert +C \end{eqnarray*} \]

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(b) To find \(\int \sec x dx\), we multiply the integrand by \(\dfrac{\sec x+\tan x}{\sec x+\tan x}.\) \[ \int \sec x dx=\int \sec x\cdot \dfrac{\sec x+\tan x}{\sec x+\tan x} dx=\int \dfrac{\sec ^{2}x+\sec x\tan x}{\sec x+\tan x} dx \]

Now the numerator equals the derivative of the denominator. So if \(u=\sec x+\tan x,\) then \(du=( \sec x\tan x+\sec ^{2}x) dx.\) \[ \int \sec x dx=\int \dfrac{du}{u}=\ln \left\vert u\right\vert +C=\ln \left\vert \sec x+\tan x\right\vert +C \]