How do you integrate $\int \frac{1}{\sqrt{e^{2 x} - 2 e^{x} - 24}} d x$ $int 1/sqrt(e^(2x)-2e^x-24)dx$ using trigonometric substitution?

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Answer 1 · 2018-06-11T11:57:07

Use the substitution $24 e^{- x} + 1 = 5 sin θ$ $24e^-x+1=5sintheta$ .

Let

$I = \int \frac{1}{\sqrt{e^{2 x} - 2 e^{x} - 24}} d x$ $I=int1/sqrt(e^(2x)-2e^x-24)dx$

Rewrite in terms of $e^{- x}$ $e^-x$ :

$I = \int \frac{e^{- x}}{\sqrt{1 - 2 e^{- x} - 24 e^{- 2 x}}} d x$ $I=inte^-x/sqrt(1-2e^-x-24e^(-2x))dx$

Complete the square in the square root:

$I = \sqrt{24} \int \frac{e^{- x}}{\sqrt{25 - {(24 e^{- x} + 1)}^{2}}} d x$ $I=sqrt24inte^-x/sqrt(25-(24e^-x+1)^2)dx$

Apply the substitution $24 e^{- x} + 1 = 5 sin θ$ $24e^-x+1=5sintheta$ :

$I = \sqrt{24} \int \frac{1}{5 cos θ} (- \frac{5}{24} cos θ d θ)$ $I=sqrt24int1/(5costheta)(-5/24costhetad theta)$

Simplify:

$I = - \frac{1}{\sqrt{24}} \int d θ$ $I=-1/sqrt24intd theta$

The integral is trivial:

$I = - \frac{1}{\sqrt{24}} θ + C$ $I=-1/sqrt24theta+C$

Reverse the substitution:

$I = - \frac{1}{\sqrt{24}} {sin}^{- 1} (\frac{24 e^{- x} + 1}{5}) + C$ $I=-1/sqrt24sin^(-1)((24e^-x+1)/5)+C$

How do you integrate ∫1√e2x−2ex−24dxint 1/sqrt(e^(2x)-2e^x-24)dx using trigonometric substitution?