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Problem 16 Easy Difficulty

Find the general indefinite integral.

$ \displaystyle \int \sec t (\sec t + \tan t)\, dt $


$\int \sec t(\sec t+\tan t) d t=\tan t+\sec t+c$

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Video Transcript

okay, We know we have this. Okay? The first thing he knows that before integrating need to simplify this to make it easier integrate. Which means we can actually distribute. So remember, when we do this but each of the terms multiple IDs on the outside, you know, of seeking squared of cheap plus seeking t times chan of g deed Thio not We know that the integral seeking squared of tea is actually simply tan of tea. And we know the integral seeking times Tangent is seeking t. So this actually makes it really simple and straightforward, if you know the shortcuts.