Ambiguity regarding the solution for $\int \sec^2x\tan x dx$

Asked Apr 19 '24 at 14:14

Active Apr 19 '24 at 14:14

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I had my first integration test today. There, I met up with a integrand that gives two different functions as a whole.

Taking $t=\tan x$:-

$$\int \sec^2x\tan x dx = \int t \ dt=\frac{t^2}{2}+C=\frac{\tan^2x}{2}+C$$

Taking $t=\sec x$:- $$\int \sec^2x\tan x dx = \int t \ dt=\frac{t^2}{2}+C=\frac{\sec^2x}{2}+C$$

Which is the correct answer to the question?

Are there any other examples except $\csc^2x\cot x$ and this one?

asked Apr 19 '24 at 14:14

Gwen

3

This (type of) question is a megaduplicate, and the answer is that both are correct, since their difference is constant. – Anne Bauval Apr 19 '24 at 14:18
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Two answer are basically the same since $\tan^2(x)+1=\sec^2(x)$, you can just change the constant of the first to get the second. – Tri Apr 19 '24 at 14:18
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Does that extra half from $\frac{\tan^2 x +1}{2}$ go away along with the $C$? @Tri – Gwen Apr 19 '24 at 14:20
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$\frac{\sec^2x}{2}+C=\frac{\tan^2x}{2}+D$, for $D=C+\frac12$. – Anne Bauval Apr 19 '24 at 14:23
@Gwen I am not sure what you meant by going away, but indefinite integral gives you a family of functions, not just one. So technically you can change C in the first to C+1/2 without any problems, if that is what you meant – Tri Apr 19 '24 at 14:23

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