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How do you integrate #int 1/sqrt(-e^(2x) +100)dx# using trigonometric substitution?

1 Answer

Jul 10, 2016

#1/10 int (dy)/cosy#

Explanation:

Making #e^{-x} = 10 cosy# we have

#- e^{-x}dx = -10 sin y dy#

and

#dx = 10 siny/e^{-x}dy = tan y dy#

so

#int (dx)/sqrt(-e^(2x) +100) equiv 1/10int (tany)/sinydy = 1/10 int (dy)/cosy#

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