Inverses Revisited

You might think that looks nothing like any of the inverse trigonometric functions, and you would be right. We are going to use our Algebra skills to make it look like something on the table of inverse trigonometric integrals. Let’s follow the recent note colored in blue and arrange the denominator accordingly. My intuition tells me the denominator can be rearranged in the form of $y=1+x^2$ because the only inverse functions with denominators in the form of $y=1-x^2$ have a square root symbol in them, which this function does not.

So we can rearrange the integral into

$-\int \frac{e^x}{(e^{2x}+4e^x+4)+1}dx$​
This helps us identify a key part of the integral that we can rearrange into something much more manageable. Notice carefully what happens.

$-\int \frac{e^x}{(e^x+2{)}^2+1}dx$​
Amazing! We can now see that instead of x we have substituted $e^x+2$, but what about the extra $e^x$ in the numerator?

Recall when we used the chain rule that we multiplied by the derivative of the inside function after taking the derivative of the outside function. The outside function in this case seems to be ${\cot }^{-1}(f(x))$ where $f(x)$, the inside function, is equal to $e^x+2$.

That’s why there is an extra $e^x$ in the numerator, because the derivative of $e^x+2$ is multiplied from the chain rule. We will get more practice with this in the u-substitution section.

We just discussed how this integral is equal to

${\mathbf{c}\mathbf{o}\mathbf{t}}^{-1}({\mathbf{e}}^{\mathbf{x}}+\mathbf{2})+\mathbf{c}$

(Remember the constant!) If you want to double check, take the derivative of this function using the table shown earlier in this section. However, by this point, I’m sure you can already use the chain rule in your head!

This is also a little tricky to solve. First let’s try to rearrange the integral into a form that includes +1 or -1 in the denominator.​
$\int \frac{1}{2\sqrt{x(1-x)}}dx$

This is the same as

$\int \frac{1}{2\sqrt{x}}\cdot \frac{1}{\sqrt{1-x}}dx$​
Can you see what we can do now...? Maybe this will help

$\int \frac{\mathbf{1}}{\mathbf{2}\sqrt{\mathbf{x}}}\cdot \frac{1}{\sqrt{1-(\sqrt{x}{)}^2}}dx$

The derivative of the inside function is highlighted in green. What’s the outside function? Looking at our table, we find that the outside function is the ${\sin }^{-1}$ function. What is the inside function though? Which function has the derivative highlighted in green?

$=\mathbf{s}\mathbf{i}{\mathbf{n}}^{-\mathbf{1}}\left(\sqrt{\mathbf{x}}\right)+\mathbf{c}$