How do you integrate #int e^x/sqrt(-e^(2x)-20e^x-96)dx# using trigonometric substitution?
1 Answer
Feb 25, 2016
Explanation:
Considering the expression under the radical sign
#-(e^x+10)=-e^(2x)-20e^x-100#
#-> -e^(2x)-20e^x-96=2^2-(e^x+10)#
Now we can see a convenient trigonometric substitution such as:
#e^x+10=2siny#
#-> e^x*dx=2cosy*dy#
#=> int e^x/sqrt(-e^(2x)-20e^x-96)dx=int (2cos y)/sqrt(4-(2siny)^2)dy#
#=int (cancel(2)cosy)/(cancel(2)sqrt(1-sin^2y))dy=int cancel(cosy)/cancel(cosy)dy=int dy=y+const.#
But
#2siny=e^x+10# =>#siny=(e^x+10)/2# =>#y=sin^(-1)((e^x+10)/2)#
Then the result is
