Question #108781

∫sec(ax) dx


1
Expert's answer
2020-04-16T17:41:38-0400

sec(ax)dxApplyu-substitution:u=xadu=adxdx=1adusec(ax)dx=sec(u)1adu=1asec(u)duUsethecommonintegral:sec(u)du=lnsec(u)+tan(u)=1a(lnsec(u)+tan(u))+CSubstitutebacku=xa=1a(lnsec(xa)+tan(xa))+C\begin{aligned} & \int{\sec }\left( ax \right)dx \\ & \\ & \text{Apply}\,\text{u-substitution}:\,u=xa \\ & \\ & \Rightarrow \,du=adx \\ & \\ & \Rightarrow \,dx=\frac{1}{a}du \\ & \\ & \int{\sec }\left( ax \right)dx=\int{\sec }\left( u \right)\cdot \frac{1}{a}du \\ & \\ & =\frac{1}{a}\int{\sec }\left( u \right)du \\ & \\ & \text{Use}\,\text{the}\,\text{common}\,\text{integral}:\quad \int{\sec }\left( u \right)du=\ln \left| \sec \left( u \right)+\tan \left( u \right) \right| \\ & \\ & =\frac{1}{a}\left( \ln \left| \sec \left( u \right)+\tan \left( u \right) \right| \right)+C \\ & \\ & \text{Substitute}\,\text{back}\,u=xa \\ & \\ & =\frac{1}{a}\left( \ln \left| \sec \left( xa \right)+\tan \left( xa \right) \right| \right)+C \\ \end{aligned}


We can use integral table given in any text book to find the well-known integral. For example, we can find the integral of sec(x) in the link below:

https://math.boisestate.edu/~wright/courses/m333/IntegralTablesStewart.pdf

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