# How do you integrate #int e^-x/(9e^(-2x)+1)^(3/2)# by trigonometric substitution?

Substituting in the integral we have

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To integrate the given expression using trigonometric substitution, we can make the substitution (u = 9e^{-2x} + 1). This implies (du = -18e^{-2x}dx), which can be simplified to (du = -\frac{18}{9e^{2x}}dx = -\frac{18}{9}du = -2du).

Therefore, the integral becomes:

[ \begin{aligned} \int \frac{e^{-x}}{(9e^{-2x}+1)^{3/2}}dx &= \int \frac{e^{-x}}{u^{3/2}} \left(-\frac{1}{2}\right)du \ &= -\frac{1}{2} \int \frac{du}{u^{3/2}} \ &= -\frac{1}{2} \int u^{-3/2} du \ &= -\frac{1}{2} \cdot \frac{u^{-1/2}}{-1/2} + C \ &= -\frac{1}{2} \cdot \frac{1}{-1/2} \cdot (9e^{-2x}+1)^{-1/2} + C \ &= 2(9e^{-2x}+1)^{-1/2} + C \ &= 2 \sqrt{9e^{-2x}+1} + C. \end{aligned} ]

So, the integral of (\frac{e^{-x}}{(9e^{-2x}+1)^{3/2}}) with respect to (x) is (2 \sqrt{9e^{-2x}+1} + C), where (C) is the constant of integration.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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