Solved: "Evaluate the following integrals. ∫dxex1−e2x"

Globokim8

Answered question

2021-10-12

Evaluate the following integrals.

\int \frac{d x}{e^{x} \sqrt{1 - e^{2 x}}}

Usamah Prosser

Skilled2021-10-13Added 86 answers

Given integral is

\int \frac{d x}{e^{x} \sqrt{1 - e^{2 x}}}

Let us substitute

u = e^{x}

\Rightarrow d u = e^{x} d x

\Rightarrow d x = e^{- x} d u

\int \frac{d x}{e^{x} \sqrt{1 - e^{2 x}}} = \int \frac{d u}{u^{2} \sqrt{1 - u^{2}}}

\int \frac{d u}{u^{2} \sqrt{1 - u^{2}}}

Now substitute

\sin v = u \Rightarrow v = \sin^{- 1} u

\cos v d v = d u

\int \frac{d u}{u^{2} \sqrt{1 - u^{2}}} = \int \frac{\cos v}{\sin^{2} v \sqrt{1 - \sin^{2} v}} d v

= \int \frac{\cos v}{\sin^{2} v \sqrt{\cos^{2} v}} d v

= \int \frac{\cos v}{\sin^{2} v (\cos v)} d v

= \int \frac{1}{\sin^{2} v} d v

= \int \frac{1}{\sin^{2} v} d v = \int \csc^{2} v d v

= - \cot v

Undoing the substitution

v = \sin^{- 1} u

\cot v = - \cot (\sin^{- 1} u)

= - \frac{\sqrt{1 - u^{2}}}{u}

Undoing the substitution

u = e^{x}

- \frac{\sqrt{1 - u^{2}}}{u} = - \frac{\sqrt{1 - e^{2 x}}}{e^{x}}

Therefore we get,

\int \frac{d x}{e^{x} \sqrt{1 - e^{2 x}}} = - \frac{\sqrt{1 - e^{2 x}}}{e^{x}} + C

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