Question

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

Answer 1

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

Explanation:

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`