Question

Anti derivative question?

Answer 1

Find f.

\[f(t) = \frac{ 4 }{ 1 + t^2}, f(0) = 1\]

Answer 2

do you mean f'(t0 is 4/(1+t^2)?

Answer 3

1/(1+t^2) reminds me of arctan function, let me double check

Answer 4

Here's my work so far (it's probably wrong):
f(t) = 4t - 4t^(-1) + c
0 = 4 - 4 + c, c = 0
f(t) = 4t - 4/t

Answer 5

The answer in the book has a weird arctan thingy... what's going on with that?

Answer 6

1/1+t^2 integrates into arctan

Answer 7

@caozeyuan Yes, the first equation should be f'(t). But f(0) = 1 (without the prime)

Answer 8

Is that something I'm supposed to memorize? How can I derive it?

Answer 9

I don't recall my teacher going over trig antiderivatives..

Answer 10

I had my calc during high school and now Im junior collge students so I dont remeber how to derive the result but it is true that if you integrate 1/1+t^2, you get arctan

Answer 11

there should be derivation online, I vaguely remember sec function or something

Answer 12

https://www.wolframalpha.com/input/?i=integrate+4%2F(1%2Bt%5E2) here is what wolfram alpha has

Answer 13

It is good to memorize that \(\arctan' (x) = \frac{1}{1+x^2}\).

Now, you must find the value of the constant \(C\) such that
\(\arctan(x)+C = 1\) when \(x=0\).

Meaning, you must solve \(\arctan(0) + C = 1\) for \(C\).

Then, construct again \(\arctan(x) + C\) with the correct \(C\) (there is only one).

Answer 14

A quick derivation of \(\frac{d}{dx} arctan(x)\):
Let
\[y = \arctan(x) \implies \tan(y) = x \]
This defines a triangle of angle \( y \) with opposite side length of \(x\) and adjacent length \(1\). The hypotenuse then has a defined side of \(\sqrt{1+x^2}\) Applying the derivative operator:
\[\frac{d}{dx}(\tan(y)) = \frac{d}{dx }(x) \implies \sec^2(y)\frac{dy}{dx} = 1 \]
Then:
\[ \frac{dy}{dx} = \frac{1}{\sec^2 y} = \cos^2 y\]
Then by definition \(\cos^2 y = \left ( \frac{1}{\sqrt{1+x^2}} \right ) ^2 \)

Then
\[ \frac{dy}{dx} = \frac{1}{1+x^2}\]