Question

Integrate the given expression, $\int{\dfrac{{{e}^{x}}}{\sqrt{{{e}^{2x}}-1}}dx}$ .

Answer 1

Hint: Assume $t={{e}^{x}}$ . Use the equation $dt={{e}^{x}}dx$ and then simplify the expression in terms of t as $\int{\dfrac{1}{\sqrt{{{t}^{2}}-1}}dt}$. Consider $t=\sec \theta$ . Use the equation $dt=\sec \theta .\tan \theta d\theta$ , and then simplify the expression $\int{\dfrac{1}{\sqrt{{{\sec }^{2}}\theta -1}}\sec \theta .\tan \theta d\theta }$ . We know that, $\int{secxdx}=ln\left| secx\left. +tanx \right| \right.$ . Use this formula and solve the expression further. Then, express $\theta$ in terms of t. And then, express t in terms of x as we assumed $t={{e}^{x}}$ .

Complete step-by-step solution -
Let us assume, $t={{e}^{x}}$
${{t}^{2}}={{e}^{2x}}$ …………….(1)
Differentiating equation (1) with respect to x, we get
$\dfrac{dt}{dx}={{e}^{x}}$
$\Rightarrow dt={{e}^{x}}dx$ ………………..(2)
Now, using equation (1) and equation (2), we can transform the expression $\int{\dfrac{{{e}^{x}}}{\sqrt{{{e}^{2x}}-1}}dx}$ .
Transforming the expression, we get $\int{\dfrac{1}{\sqrt{{{t}^{2}}-1}}dt}$ ……….(3)
Now, let us assume,
$t=\sec \theta$ …………..(4)
Differentiating equation (4) with respect to $\theta$ , we get
$\dfrac{dt}{d\theta }=\sec \theta .\tan \theta$
$\Rightarrow dt=\sec \theta .\tan \theta d\theta$ ………………..(5)
Now, using equation (5), we can transform equation (3)
$\int{\dfrac{1}{\sqrt{{{t}^{2}}-1}}dt}$
$=\int{\dfrac{1}{\sqrt{{{\sec }^{2}}\theta -1}}\sec \theta .\tan \theta d\theta }$ ……………….(6)
We know the identity, ${{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1$
$\Rightarrow {{\sec }^{2}}\theta -1={{\tan }^{2}}\theta$ ………………………(7)
Now, using equation (7), we can transform equation (6)
$=\int{\dfrac{1}{\sqrt{{{\sec }^{2}}\theta -1}}\sec \theta .\tan \theta d\theta }$

& =\int{\dfrac{1}{\sqrt{{{\tan }^{2}}\theta }}\sec \theta .\tan \theta d\theta } \\\ & =\int{\dfrac{1}{\tan \theta }}.\sec \theta \tan \theta d\theta \\\ \end{aligned}$$ $$=\int{\sec \theta d\theta }$$ …………………(8) We know that, $$\int{secxdx}=ln\left| secx\left. +tanx \right| \right.$$ ………….(9) Replacing x by $$\theta $$ in the equation (9), we get $$\int{secxdx}=ln\left| secx\left. +tanx \right| \right.$$ $$\Rightarrow \int{\sec \theta d\theta }=\ln \left| \sec \theta \left. +\tan \theta \right| \right.$$ …………….(10) From equation (4), we have $$t=\sec \theta $$ We can find the value of $$\tan \theta $$ using equation (7). $$\begin{aligned} & {{\sec }^{2}}\theta -1={{\tan }^{2}}\theta \\\ & \Rightarrow {{t}^{2}}-1={{\tan }^{2}}\theta \\\ & \Rightarrow \tan \theta =\sqrt{{{t}^{2}}-1} \\\ \end{aligned}$$ Initially, we have assumed $$t={{e}^{x}}$$. Now, putting the value of t, we get $$t=\sec \theta ={{e}^{x}}$$ $$\begin{aligned} & \tan \theta =\sqrt{{{t}^{2}}-1} \\\ & \Rightarrow \tan \theta =\sqrt{{{e}^{2x}}-1} \\\ \end{aligned}$$ Now, putting the values of $$\sec \theta $$ and $$\tan \theta $$ in equation (10), we get $$\Rightarrow \int{\sec \theta d\theta }=\ln \left| \sec \theta \left. +\tan \theta \right| \right.$$ $$=\ln \left| {{e}^{x}}\left. +\sqrt{{{e}^{2x}}-1} \right| \right.$$ Therefore, $$\int{\dfrac{{{e}^{x}}}{\sqrt{{{e}^{2x}}-1}}dx}=\ln \left| {{e}^{x}}\left. +\sqrt{{{e}^{2x}}-1} \right| \right.$$ . Note: In this question, one can think to assume $${{e}^{2x}}$$ as t in the expression $$\int{\dfrac{{{e}^{x}}}{\sqrt{{{e}^{2x}}-1}}dx}$$ . But, if we do so then our equation will look like $$\int{\sqrt{t}.\dfrac{1}{\sqrt{{{t}^{2}}-1}}dx}$$ . Here, we are unable to replace $$dx$$ as $$dt$$ . We can see that this expression has become complex to be solved further. So, this approach is not suitable for the integration of this expression.