Understanding the Integration of Inverse Trigonometric Functions

[escryan]

I'm just curious as to how

$$\int (\frac{1}{1+x^{2}}) dx$$

comes to be

$$\tan^{-1} (x)$$


I was able to find the formula on a table of integrals, but I'd just like to know why it works like this, and why we can't use a natural log rule or a substitution method to find this out.

Thanks for reading!

 

[rocomath]

$$y=\tan^{-1}x$$

$$x=\tan y$$ (Implicit Differentiation)

$$y'=\frac{1}{\sec^2 y}$$

Substitute with a trig identity and you will see why it's true.

 

[RyanSchw]

You should surely be able to see why using u-substitution with the natural log will not work. My guess is that you are just given that anything in that form is going to be arctan and you will learn how to do this later on with Trigonometric Substitutions.

Remember

$$Sin^2(x) + Cos^2(x) = 1$$

and from that

$$tan^2(x) + 1 = Sec^2(x)$$

You can these use these identities when you notice that $1+x^2$ in

$$\int \frac{1}{1+x^2}$$

can be substituted as $sec^2(x)$

there is obviously more to it than that, but to get you started.