Proving a standard integral? (why the integral of 1 / (1 + x^2) is equal to arctan x + c)

[Skelly4444]

In mathematics, if we're asked to prove a standard integral, where can we find step by step solutions, or are we just supposed to accept it as it stands?

I am trying to find out exactly why the integral of 1 / (1 + x^2) is equal to arctan x + c

Any advice would be greatly appreciated

 

[MarkFL]

It would be sufficient to show that:

\(\displaystyle \frac{d}{dx}(\arctan(x))=\frac{1}{x^2+1}\)

Let's begin with:

\(\displaystyle y=\arctan(x)\)

Or:

\(\displaystyle x=\tan(y)\)

Differentiate with respect to x...what do you find?

 

[Skelly4444]

Many thanks for replying.

I have had another look at this using the substitution x=tan u and I managed to integrate it and obtain the answer of arctan x in the end

It wasn't as difficult as I'd imagined it would be.

 

[MarkFL]

Both methods rely on a Pythagorean identity. ?

 

[MarkFL]

It would be sufficient to show that:

\(\displaystyle \frac{d}{dx}(\arctan(x))=\frac{1}{x^2+1}\)

Let's begin with:

\(\displaystyle y=\arctan(x)\)

Or:

\(\displaystyle x=\tan(y)\)

Differentiate with respect to x...what do you find?

We get:

\(\displaystyle 1=\sec^2(y)\frac{dy}{dx}\)

\(\displaystyle \frac{1}{\sec^2(y)}=\frac{dy}{dx}\)

Apply a Pythagorean identity:

\(\displaystyle \frac{1}{\tan^2(y)+1}=\frac{dy}{dx}\)

\(\displaystyle \frac{1}{x^2+1}=\frac{dy}{dx}\)

Shown as desired.

 

[MarkFL]

You likely began with:

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

Let:

\(\displaystyle x=\tan(u)\rightarrow\,dx=\sec^2(u)\,du\)

\(\displaystyle \int\frac{\sec^2(u)}{\tan^2(u)+1}\,du\)

Applying that same Pythagorean identity:

\(\displaystyle \int\,du=u+C=\arctan(x)+C\)

 

[Steven G]

In mathematics, if we're asked to prove a standard integral, where can we find step by step solutions, or are we just supposed to accept it as it stands?

Never accept any mathematics.