Suppose you have a function, f(x) = sinx*cos³x and you want its integral.

One way to do it is u = sin x, then du = cosx dx.

This way you would get: 1/2sin²x – 1/4sin^4x +C.

Another way is, u = cos x, then du = -sinxdx.

This way you get: -1/4cos^4x + C out.

How come these are both correct?

(Or I must have made a serious mistake somewhere…)

## Answer

It is indeed possible that the result of an indefinite integral looks different. The solution of one __indefinite__ after all, integral is not one function, but infinitely many functions that are mutually equal on a constant. We therefore write the solution as a single function + constant. What that one function, the representative, (a kind of “ambassador”) of the solution is, will depend on the method followed. Especially with trigonometric solutions and with logarithms you can sometimes get two very different representatives, which, however, are equal except for one constant.

For example, take the integral of sin(x).cos(x)

You can then get whatever you want, depending on whether you put the sine or the cosine in the differential dx as in your example:

integral = 0.5 sin^{2}(x) +K

integral = – 0.5 cos^{2}(x) + K

however, with cos^{2}(x) + sin^{2}(x) = 1 you can convert one into the other. The extra term 1 disappears in K

But you could also first say: sin(x).cos(x) = 0.5 sin(2x), so that then:

integral : – 0.25 cos(2x) + K

it is yet another representative, which via trigonometry (double angle formulas) is again equal to the other two except for a constant.

If you have integrals where the solution contains logarithms and trigonometry, the differences can sometimes be very large, but they are only apparent differences, as in the examples above.

Specifically for your specific integral: simply with the basic formula cos^{2}(x)+sin^{2}(x) = 1 you can show that the two representatives are equal except for a constant.

## Answered by

#### prof.dr. Paul Hellings

Department of Mathematics, Fac. IIW, KU Leuven

Old Market 13 3000 Leuven

https://www.kuleuven.be/

.