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

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Answer 1 · 2018-04-09T05:54:54

$color(red)(-sqrt(-e^(2x)-25)+C$

First factor out $i$ from the root.

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

$inte^(2x)/sqrt(-1(e^(2x)+25))dx$

$inte^(2x)/(sqrt(-1)sqrt(e^(2x)+25))dx$

$1/iinte^(2x)/sqrt(e^(2x)+25)dx$

$-iinte^(2x)/sqrt(e^(2x)+25)dx$

Now we can perform trig substitution.

It's important to recognize:

$e^(2x)=e^x*e^x$

Let

$tantheta=e^x/5$

$e^x=5tantheta$

$e^x*dx=5sec^2theta* d theta$

$sectheta=sqrt(e^(2x)+25)/5$

$5sectheta=sqrt(e^(2x)+25)$

Substituting, we get:

$-iint(5tantheta)/(5sectheta)5sec^2theta *d theta$

$-iint5tanthetasectheta*d theta$

Integrating gives us:

$-i[5sectheta]+C$

$-i[sqrt(e^(2x)+25)]+C$

$color(red)(-sqrt(-e^(2x)-25)+C$

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