Indefinite integral &#x222B;<!-- ∫ --> 1 2 &#x2212;<!-- − -->

Noelle Wright

Noelle Wright

Answered question

2022-05-10

Indefinite integral 1 2 cos ( x ) d x has discontinuities. How to fix?
Using the standard tangent half-angle substitution, we get
1 2 cos ( x ) d x = 2 3 tan 1 ( 3 tan x 2 ) + C
The resulting antiderivatives are piecewise continuous functions with discontinuities at x = ( 2 k + 1 ) π. However, the integrand is continuous everywhere, so any antiderivative must be continuous everywhere (??) according to FTC. So, how do we deal with this problem? What is the correct form of the antiderivatives? Is there a form that avoids the discontinuities?

Answer & Explanation

Superina0xb4i

Superina0xb4i

Beginner2022-05-11Added 17 answers

Step 1
Although I'm really late with the answer, here's something other people might find interesting.
There is a continous elementary antiderivative,
F ( x ) = 2 3 ( x 2 + tan 1 ( ( 3 1 ) tan x 2 1 + 3 tan 2 x 2 ) )
but finding it may be somewhat inconvenient. Here's the how.
First of all, we carefully look at the function tan 1 ( tan x ) . Quite interesting plot, isn't it?
So, we see (check it!) that the floor function can be written as
f l o o r ( x ) = 2 x 1 2 1 π tan 1 ( tan ( 2 x 1 2 π ) )
Next, we find the precise floor expression that eliminates the discontinuities in your antiderivative. That part of the work is already done in one of the answers,
1 2 ( x π + 1 ) = 1 π ( x 2 tan 1 ( tan ( x 2 ) ) )
Step 2
Then you add this staircase function to your antiderivative, and end up with
F ( x ) = 2 3 [ tan 1 ( 3 tan x 2 ) + x 2 tan 1 ( tan x 2 ) ]
which is already cool if you don't mind stuff like tan 1 ( tan ( x ) ). But since we can combine inverse tangent sums by using the formula tan 1 ( a ) + tan 1 ( b ) = tan 1 ( a + b 1 a b ) , we can go even further!
Combining inverse tangents and simplifying the resulting expression, we finally get a fully continuous antiderivative F that we can use to calculate definite integrals without paying attention to the limits.
lurtzslikgtgjd

lurtzslikgtgjd

Beginner2022-05-12Added 3 answers

Step 1
Suppose you want to it from x = 0 to 2 π using tan ( x / 2 ) = t. A transformation t(x) needs to be continuous and monotonic (without max/min). But ( x ) = tan ( x / 2 ) is discontinuous at x = π, so it is a bad substitution. A bad substitution can be made to work but breaking the domain of integration at the discontinuities.
One way is I = 0 2 π d x 2 cos x = 0 + π d x 2 cos x + π + 2 π d x 2 cos x .
Step 2
This way tan ( x / 2 ) | x = π = + and tan ( x / 2 ) | π + = .. So you get I = 2 3 [ ( π 2 0 ) + ( 0 π 2 ) ] = 2 π 3 ..
Else, the other way is
You may directly use the domain reduction by 0 2 a f ( x ) d x = 2 0 a f ( x ) d x ,   i f   f ( 2 a x ) = f ( x ) .
You get I = 2 0 π d x 2 cos x .

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?