Where is the absolute value when computing antiderivatives? Here is a typical second-semester singl

Yahir Tucker

Yahir Tucker

Answered question

2022-06-26

Where is the absolute value when computing antiderivatives?
Here is a typical second-semester single-variable calculus question:
1 1 x 2 d x
Students are probably taught to just memorize the result of this since the derivative of arcsin(x) is taught as a rule to memorize. However, if we were to actually try and find an antiderivative, we might let x = sin θ d x = cos θ d θ
so the integral may be rewritten as cos θ 1 sin 2 θ d θ = cos θ cos 2 θ d θ
At this point, students then simplify the denominator to just cos θ, which boils the integral down to
1 d θ = θ + C = arcsin x + C
which is the correct antiderivative. However, by definition, x 2 = | x | , implying that the integral above should really be simplified to cos θ | cos θ | d θ = ± 1 d θ depending on the interval for θ. At this point, it looks like the answer that we will eventually arrive at is different from what we know the correct answer to be.

Answer & Explanation

Aiden Norman

Aiden Norman

Beginner2022-06-27Added 16 answers

Step 1
This is a good question. Note that the domain for the arcsine function, arcsin ( x ) = 0 x 1 1 t 2 d t, is | x | 1.
Hence, upon enforcing the substitution t = sin ( θ ), then for | θ | π / 2, cos ( θ ) 0.
Step 2
We could have chosen alternatively θ ( π / 2 + n π , π / 2 + n π ) from which we see that sgn ( cos ( θ ) ) = { 1 , n even 1 , n odd
For odd values of n, the substitution x = sin ( θ ) yields
1 1 x 2 d x = θ + C = arcsin ( x ) + C = n π arcsin ( x ) + C = arcsin ( x ) + C
And hence, the antiderivative 1 1 x 2 d x = arcsin ( x ) + C is valid regardless of whether n is even or odd.
Cory Patrick

Cory Patrick

Beginner2022-06-28Added 6 answers

Explanation:
By the FTC, the answer is F ( x ) = 0 x 1 1 t 2 d t + C for | x | < 1.. But can we write F(x) in terms of functions that we understand better? Yes, in this case, by using the basic substitution theorem, which is usually lost in the haze by teaching assistants, textbook writers, and too many professors as well. The real answer to your question lies in understanding this theorem thoroughly.

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