How do you integrate #int e^(2x)/sqrt(-e^(2x) +25)dx# using trigonometric substitution?

Question

Evelyn Agard · Accepted Answer

#=-5cos(sin^-1(e^x/5))+C#

#inte^(2x)/sqrt(-e^(2x)+25)dx#

apply trigonometric substitution

#5sintheta=e^x# #5costheta*d(theta)=e^xdx#

#inte^(2x)/sqrt(-e^(2x)+25)dx#=#int(5sintheta*5costheta)/sqrt(25-25sin^2theta)d(theta)#

#25-25sin^2theta=25cos^2theta#

simplify,

#int(5sintheta*5costheta)/sqrt(25-25sin^2theta)d(theta)#=#5intsinthetad(theta)#

#=-5costheta+C#

reverse the trigonometric substitution

#theta=sin^-1(e^x/5)#

so your answer will be:

#=-5cos(sin^-1(e^x/5))+C#

Ezekiel Alexander · Answer

undefined To integrate $\int \frac{e^{2x}}{\sqrt{-e^{2x} + 25}} \, dx$ using trigonometric substitution, let $u = -e^{x}$. Then, $du = -e^{x} \, dx$. Rewrite the integral:

$$
\begin{align*}
\int \frac{e^{2x}}{\sqrt{-e^{2x} + 25}} \, dx &= \int \frac{-1}{\sqrt{u^2 - 25}} \, du \
&= -\int \frac{1}{\sqrt{u^2 - 25}} \, du
\end{align*}
$$

Now, let $u = 5\sec(\theta)$. Then, $du = 5\sec(\theta)\tan(\theta) \, d\theta$. Rewrite the integral:

$$
\begin{align*}
-\int \frac{1}{\sqrt{u^2 - 25}} \, du &= -\int \frac{1}{\sqrt{25\sec^2(\theta) - 25}} \cdot 5\sec(\theta)\tan(\theta) \, d\theta \
&= -\int \frac{1}{\sqrt{25(\sec^2(\theta) - 1)}} \cdot 5\sec(\theta)\tan(\theta) \, d\theta \
&= -\int \frac{1}{\sqrt{25\tan^2(\theta)}} \cdot 5\sec(\theta)\tan(\theta) \, d\theta \
&= -\int \frac{1}{5\tan(\theta)} \cdot 5\sec(\theta)\tan(\theta) \, d\theta \
&= -\int \sec(\theta) \, d\theta \
&= -\ln|\sec(\theta) + \tan(\theta)| + C
\end{align*}
$$

Finally, revert back to the variable $x$ using the original substitution $u = -e^{x}$:

$$
-\ln|\sec(\theta) + \tan(\theta)| + C = -\ln|-\frac{1}{e^{x}} + \frac{\sqrt{-e^{2x} + 25}}{e^{x}}| + C
$$

$$
= \ln|\frac{\sqrt{-e^{2x} + 25} - 1}{e^{x}}| + C
$$

This is the final answer.