How do you integrate #int 1/sqrt(-e^(2x) +81)dx# using trigonometric substitution?

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Answer 1 · 2016-01-12T10:57:09

#int 1/sqrt(-e^(2x) + 81)# = #1/9 ln(9-sqrt(81-e^(2x)))-x/9+C#

Use the Trigonometric substitution

#e^x = 9* sin theta#

#e^x# #dx# = #9* cos theta # #d##theta#

#e^(2x)= 81 sin^2 theta#

#dx#=#(9cos theta)/(9sin theta# #d##theta#

#dx#=#cos theta/sin theta# #d##theta#

#int 1/sqrt(81-e^(2x))##dx#

#int 1/sqrt(81-81 sin ^2 theta)##cos theta/sin theta# #d##theta#

#int 1/(sqrt(81)*sqrt(1-sin ^2 theta))##cos theta/sin theta# #d##theta#

#int 1/(9sqrt(cos ^2 theta))##cos theta/sin theta# #d##theta#

#1/9* int csc theta # #d##theta#

#1/9*ln(csc theta - cot theta)+C#

#1/9*ln(9/e^x-sqrt(81-e^(2x))/e^x)+C#

#1/9*ln((9-sqrt(81-e^(2x)))/e^x)+C#

#1/9*(ln(9-sqrt(81-e^(2x)))-ln e^x)+C#

#1/9*(ln(9-sqrt(81-e^(2x)))-x)+C#

#1/9*ln(9-sqrt(81-e^(2x)))-x/9+C#