FANDOM


$ \int\sec(x)dx=\ln\bigl(|\sec(x)+\tan(x)|\bigr)+C $

Proof

$ \int\sec(x)dx=\int\dfrac{\sec(x)[\sec(x)+\tan(x)]}{\sec(x)+\tan(x)}\,dx $

Let $ u=\sec(x)+\tan(x) $

$ \begin{align}\frac{du}{dx}&=\sec(x)\tan(x)+\sec^2(x)\\ &=\sec(x)[\tan(x)+\sec(x)]\end{align} $
$ \begin{align}\therefore\int\frac{\sec(x)[\sec(x)+\tan(x)]}{\sec(x)+\tan(x)}\,dx&=\int\frac{du}{u}\\&=\ln\big(|u|\big)+C\\&=\ln\big(|\sec(x)+\tan(x)|\big)+C \end{align} $
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