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

$e^x = 9* sin theta$

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

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

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

$dx$ = $cos theta/sin theta$ $d$ $theta$

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

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

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

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

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

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

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

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

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

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

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

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