Question

Evaluate the integral\(\int {\frac{{{\rm{dx}}}}{{{{\rm{e}}^{\rm{x}}}\sqrt {{\rm{1 - }}{{\rm{e}}^{{\rm{ - 2x}}}}} }}} \).

Short answer

The integral value of the given equation is\(\int {\frac{{{\rm{dx}}}}{{{{\rm{e}}^{\rm{x}}}\sqrt {{\rm{1 - }}{{\rm{e}}^{{\rm{ - 2x}}}}} }}} {\rm{ = - si}}{{\rm{n}}^{{\rm{ - 1}}}}\left( {{{\rm{e}}^{{\rm{ - x}}}}} \right){\rm{ + C}}\).

Step-by-step solution

Expand the equation.

\({\rm{I = }}\int {\frac{{{\rm{dx}}}}{{{{\rm{e}}^{\rm{x}}}\sqrt {{\rm{1 - }}{{\rm{e}}^{{\rm{ - 2x}}}}} }}} \)

Let

\(\begin{aligned}{c}{\rm{ u = }}{{\rm{e}}^{{\rm{ - x}}}} \to {\rm{du}}\\{\rm{ = - }}{{\rm{e}}^{{\rm{ - x}}}}{\rm{dx}} \to {\rm{dx}}\\{\rm{ = - }}\frac{{{\rm{du}}}}{{{{\rm{e}}^{{\rm{ - x}}}}}}\\{\rm{ = - }}\frac{{{\rm{du}}}}{{\rm{u}}}\end{aligned}\)

\({\rm{I = - }}\int {\frac{{{\rm{du}}}}{{\sqrt {{\rm{1 - }}{{\rm{u}}^{\rm{2}}}} }}} \)

Evaluate the equation.

Known value\(\int {\frac{{{\rm{dx}}}}{{\sqrt {{\rm{1 - }}{{\rm{x}}^{\rm{2}}}} }}} {\rm{ = si}}{{\rm{n}}^{{\rm{ - 1}}}}{\rm{x + C}}\)

\({\rm{I = - si}}{{\rm{n}}^{{\rm{ - 1}}}}{\rm{(u) + C}}\)

Substitute \({\rm{u = }}{{\rm{e}}^{{\rm{ - x}}}}\)

\({\rm{I = - si}}{{\rm{n}}^{{\rm{ - 1}}}}\left( {{{\rm{e}}^{{\rm{ - x}}}}} \right){\rm{ + C}}\)

Therefore, the integral value of the given equation is\(\int {\frac{{{\rm{dx}}}}{{{{\rm{e}}^{\rm{x}}}\sqrt {{\rm{1 - }}{{\rm{e}}^{{\rm{ - 2x}}}}} }}} {\rm{ = - si}}{{\rm{n}}^{{\rm{ - 1}}}}\left( {{{\rm{e}}^{{\rm{ - x}}}}} \right){\rm{ + C}}\).