Question

@cebroski
I don't know if you were interested an comparing antiderivatives or not but you can come here if you do... Like pretend we want to integrate sin(2x) w.r.t x

Answer 1

there are a couple of ways I can think to integrate it
and we will get different looking results

Answer 2

but you can still manipulate them to show they are the same answer

Answer 3

\[\int\limits_{}^{}\sin(2x) dx=\frac{-1}{2}\cos(2x)+C \\ \int\limits_{}^{}\sin(2x) dx=\int\limits_{}^{}2\sin(x)\cos(x)dx\\ =\int\limits_{}^{}2u du=u^2+C=\sin^2(x)+C \text{ where } u=\sin(x) \\ \int\limits_{}^{}\sin(2x)dx=\int\limits_{}^{}2\sin(x)\cos(x) dx\\ =\int\limits_{}^{}-2udu=-u^2+C=-\cos^2(x)+C \text{ where } u =\cos(x)\]

Answer 4

but we can show the first answer can be written as the second answer or even the third answer

Answer 5

\[\frac{-1}{2}\cos(2x) \\ =\frac{-1}{2}(\cos^2(x)-\sin^2(x)) \text{ double angle identity for cos } \\ =\frac{-1}{2}(1-\sin^2(x)-\sin^2(x)) \text{ by pythagorean identity } \\ =\frac{-1}{2}(1-2\sin^2(x)) \\ =\frac{-1}{2}+\sin^2(x) \text{ second answer } \\ =\frac{-1}{2} +1-\cos^2(x) \text{ by pythagorean identity } \\ =\frac{1}{2}-\cos^2(x) \text{ third answer }\]

Answer 6

here is an algebraic one...
\[\int\limits_{}^{}x^2dx=\frac{x^3}{3}+C \\ \int\limits_{}^{}xxdx=\frac{x^2}{2}x-\int\limits_{}^{}\frac{x^2}{2}(1) dx =\frac{x^2}{2}-\frac{x^3}{6}+C \]
but I know this one isn't as weird because the x^3/3 and the x^2/2-x^3/6 don't differ by a constant

Answer 7

I mean they differ by the constant 0 (lol)

Answer 8

nice :) here is another similar trig one
\[ f(x) = \tan x \sec^2x \]

\(u = \tan x \implies \int f(x) dx = \frac{\tan^2x}{2}+C\)

\(u = \sec x \implies \int f(x) dx = \frac{\sec^2x}{2}+C\)

Answer 9

\[\sin^2(x)+\cos^2(x)=1 \\ \tan^2(x)+1=\sec^2(x) \\ \]
so we could rewrite @ganeshie8 's first answer as his second answer using Pythagorean identities again
\[\frac{\tan^2(x)}{2} \\ =\frac{\sec^2(x)-1}{2} \text{ by pythagoeren identity }\\ =\frac{\sec^2(x)}{2}-\frac{1}{2}\]

Answer 10

What about this:
\[\text{ Let } y=\frac{x-1}{x+1} \\ \text{ by quotient rule we have } y'=\frac{(x+1)-(x-1)}{(x+1)^2}=\frac{2}{(x+1)^2} \\ \text{ but when we try \to integrate y' we get: } \\ \int\limits_{}^{}y' dx=\int\limits_{}^{}\frac{2}{(x+1)^2}dx= \frac{2(x+1)^{-2+1}}{-2+1}+C =\frac{2(x+1)^{-1}}{-1}+C =\frac{-2}{x+1}+C \\ \text{ and y was } y=\frac{x-1}{x+1}=\frac{x+1-2}{x+1}=1-\frac{2}{x+1}=1+\frac{-2}{x+1}\]

Answer 11

http://tutorial.math.lamar.edu/Classes/CalcI/ConstantOfIntegration.aspx
also here is paul's notes talking about the same thing