MHB Trigonometric Integral, weird results

[Yankel]

Hello all,

I am trying to solve the integral:

\[\int cot(x)\cdot csc^{2}(x)\cdot dx\]

If I use a substitution of u=cot(x), I get

\[-\frac{1}{2}cot^{2}(x)+C\]

which is the correct answer in the book, however, if I do this:

\[\int \frac{cos(x)}{sin^{3}(x)}dx\]

I get, using a substitution u=sin(x)

\[-\frac{1}{2}csc^{2}(x)+C\]

which according to MAPLE, is also correct. In addition, according to MAPLE the move I made for the new function, is also correct !

I am confused !

 

[Opalg]

Those two answers only differ by a constant, because $\cot^2 x+ 1 = \csc^2x$.

 

[Yankel]

Mystery solved then...I never thought of that...nice ! ;-)

 

[MarkFL]

Mystery solved then...I never thought of that...nice ! ;-)

It's a "rite of passage" for calculus students. (Smile)