How do you integrate #int 1/sqrt(e^(2x)+12e^x-45)dx# using trigonometric substitution?

2 Answers
Jan 13, 2016

#int frac{dx}{sqrt{e^{2x}+12e^x-45}} = frac{sqrt5*arcsin(frac{2}{3}-5e^{-x})}{15}+C#,
where #C# is the constant of integration.

Explanation:

Completing the square at the denominator gives

#e^{2x} + 12e^x - 45 -= (e^x + 6)^2 - 81#

To make use of the identity

#sec^2u - 1 -= tan^2u#,

substitute #9secu = e^x + 6#.

Jan 13, 2016

nice.. ...and the mighty Pythagorean right triangle supports this....
enter image source here

From this tiny little triangle you can glean TONS of valuable information and insight. Simple geometries can typically yield a beautiful perspective of things. On your own and using this triangle as your reference, find:
# 1) tan(theta)#
# 2) sec(theta)#
# 3) sin(theta)#
# 4) x#
# 5) dx/(d theta)#
# 6) dx#
# 7) sqrt(e^(2x)+12e^x-45)#

No need to memorize formulas......that is WAY too much work and expended effort!!

Explanation:

Ok it work my bad