Indefinite integrals with different solutions?

[AntSC]

Indefinite integrals with different solutions?

Homework Statement

\int \csc ^{2}2x\cot 2x\: dx
Solve without substitution using pattern recognition

Homework Equations

As above

The Attempt at a Solution

To try a function that, when differentiated, is of the same form as the integrand.
The two solutions are attached.
My question is that both functions i tried, differentiated to the integrand, but as a result they yield a different solution to the integral. These two functions look very similar. I don't understand what the significance of this is. I've seen a similar result when playing around with other integrals. Any light on this would be really helpful. Cheers.

 

[Dick]

Homework Statement

\int \csc ^{2}2x\cot 2x\: dx
Solve without substitution using pattern recognition

Homework Equations

As above

The Attempt at a Solution

To try a function that, when differentiated, is of the same form as the integrand.
The two solutions are attached.
My question is that both functions i tried, differentiated to the integrand, but as a result they yield a different solution to the integral. These two functions look very similar. I don't understand what the significance of this is. I've seen a similar result when playing around with other integrals. Any light on this would be really helpful. Cheers.

cosec^2=cot^2+1. Two functions that differ by a constant have the same derivative - so both are fine forms for the indefinite integral.

 

[arildno]

Remember that:
\frac{\cos^{2}(y)}{\sin^{2}(y)}=\frac{1}{\sin^{2}(y)}-1
due to the age-oldest relation between cos and sin.

 

[AntSC]

Aha! Of course! Totally forgot that. I was looking at differences in constants but completely missed the identity.
So in general, if solutions to an integral are different (if one is on the ball to spot it) then the difference is due to the constants?

 

[arildno]

As long as your purported antiderivatives are TRUE antiderivatives, then they do only differ by at most a non-zero constant.

 

[AntSC]

Got it. Thanks very much :)