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Question: Integrate \(\dfrac{{dx}}{{{{\left( {1 - {e^{2x}}} \right)}^{\dfrac{1}{2}}}}}\)....

Integrate dx(1−e2x)12\dfrac{{dx}}{{{{\left( {1 - {e^{2x}}} \right)}^{\dfrac{1}{2}}}}}.

Explanation

Solution

First assume 1−e2x=t21 - {e^{2x}} = {t^2}. Then transform dxdx in terms of dθd\theta . After that change the original equation in the form of tt. Then factor the denominator and apply partial fraction decomposition on the term. After that integrate the terms and apply log properties to get the desired result.

Complete step by step answer:
Let us assume the given integral as
⇒I=∫11−e2xdx\Rightarrow I = \int {\dfrac{1}{{\sqrt {1 - {e^{2x}}} }}dx}
Let us assume 1−e2x=t21 - {e^{2x}} = {t^2}.
Then, differentiate the function to get the value of dtdt,
⇒−2e2xdx=2tdt\Rightarrow - 2{e^{2x}}dx = 2tdt
Cancel out the common factors,
⇒tdt=−e2xdx\Rightarrow tdt = - {e^{2x}}dx ….. (1)
Now find the value of e2x{e^{2x}} from the assumption,
⇒1−e2x=t2\Rightarrow 1 - {e^{2x}} = {t^2}
Move 1 to another part,
⇒−e2x=t2−1\Rightarrow - {e^{2x}} = {t^2} - 1
Substitute the value in equation (1),
⇒tdt=(t2−1)dx\Rightarrow tdt = \left( {{t^2} - 1} \right)dx
Divide both sides by (t2−1)\left( {{t^2} - 1} \right),
⇒dx=tdt(t2−1)\Rightarrow dx = \dfrac{{tdt}}{{\left( {{t^2} - 1} \right)}}
The new equation will be,
⇒I=∫1t2×tt2−1dt\Rightarrow I = \int {\dfrac{1}{{\sqrt {{t^2}} }} \times \dfrac{t}{{{t^2} - 1}}dt}
Simplify the terms,
⇒I=∫1t×tt2−1dt\Rightarrow I = \int {\dfrac{1}{t} \times \dfrac{t}{{{t^2} - 1}}dt}
Cancel out the common factors,
⇒I=∫1t2−1dt\Rightarrow I = \int {\dfrac{1}{{{t^2} - 1}}dt}
Factor the denominator,
⇒I=∫1(t+1)(t−1)dt\Rightarrow I = \int {\dfrac{1}{{\left( {t + 1} \right)\left( {t - 1} \right)}}dt}
Now apply the partial fraction decomposition,
⇒I=∫(12×1t−1−12×1t+1)dt\Rightarrow I = \int {\left( {\dfrac{1}{2} \times \dfrac{1}{{t - 1}} - \dfrac{1}{2} \times \dfrac{1}{{t + 1}}} \right)dt}
Break the terms into two parts,
⇒I=∫12×1t−1dt−∫12×1t+1dt\Rightarrow I = \int {\dfrac{1}{2} \times \dfrac{1}{{t - 1}}dt} - \int {\dfrac{1}{2} \times \dfrac{1}{{t + 1}}dt}
Now, we know that,
∫af(x)dx=a∫f(x)dx\int {af\left( x \right)dx} = a\int {f\left( x \right)dx}
Apply on the above equation,
⇒I=12∫1t−1dt−12∫1t+1dt\Rightarrow I = \dfrac{1}{2}\int {\dfrac{1}{{t - 1}}dt} - \dfrac{1}{2}\int {\dfrac{1}{{t + 1}}dt}
Now integrate the terms,
⇒I=12ln⁡(t−1)−12ln⁡(t+1)\Rightarrow I = \dfrac{1}{2}\ln \left( {t - 1} \right) - \dfrac{1}{2}\ln \left( {t + 1} \right)
Now, we know that,
ln⁡x−ln⁡y=ln⁡xy\ln x - \ln y = \ln \dfrac{x}{y}
Using the above property,
⇒I=12ln⁡t−1t+1\Rightarrow I = \dfrac{1}{2}\ln \dfrac{{t - 1}}{{t + 1}}
Put back the value t=1−e2xt = \sqrt {1 - {e^{2x}}} in the above integral I, we get
∴I=12ln⁡1−e2x−11−e2x+1\therefore I = \dfrac{1}{2}\ln \dfrac{{\sqrt {1 - {e^{2x}}} - 1}}{{\sqrt {1 - {e^{2x}}} + 1}}

Hence, the value of dx(1−e2x)12\dfrac{{dx}}{{{{\left( {1 - {e^{2x}}} \right)}^{\dfrac{1}{2}}}}} is 12ln⁡1−e2x−11−e2x+1\dfrac{1}{2}\ln \dfrac{{\sqrt {1 - {e^{2x}}} - 1}}{{\sqrt {1 - {e^{2x}}} + 1}}.

Note: Differentiation and integration are the two important concepts of calculus. Calculus is a branch of mathematics that deals with the study of problems involving a continuous change in the values of quantities. Differentiation refers to simplifying a complex function into simpler functions. Integration generally refers to summing up the smaller function to form a bigger unit.
Indefinite integrals are those integrals that do not have any limit of integration. It has an arbitrary constant. Definite integrals are those integrals which have an upper and lower limit. Definite integral has two different values for the upper limit and lower limit when they are evaluated. The final value of a definite integral is the value of integral to the upper limit minus the value of the definite integral for the lower limit.