U-Substitution in trig integral

[ago01]

Homework Statement

$\int \frac{\csc{x}\cot{x}}{1+\csc^2{x}}dx$

Relevant Equations

U-Sub

$\int \frac{\csc{x}\cot{x}}{1+\csc^2{x}}dx$

Let $u = \csc{x}$

then

$-du = \csc{x}\cot{x}dx$

So,

$\int \frac{\csc{x}\cot{x}}{1+\csc^2{x}}dx$

$-\int \frac{1}{1+u^2}du = -\arctan{u} + C$

$-\arctan{\csc{x}} + C$

This answer was wrong. The actual answer involved fully simplifying the fraction and using

$u = \sin{x}$.

I understand how this approach is done and it makes complete sense.

What I'm confused about is where I went wrong in my derivation. I don't see my mistake, and the only difference is I ended up with $\csc{x}$ and a negative sign.

Could you help me spot my error? Usually bad u-subs makes things harder so I must've made a dumb subtle mistake I can't see.

 

[fresh_42]

I suspect that the mistake is hidden in the boundaries. $\tan^{-1}(\sin(x))+\tan^{-1}(\csc(x))$ is a step function:
https://www.wolframalpha.com/input?i=arctan(sin(x))+++arctan(1/sin(x))=

 

[ago01]

I suspect that the mistake is hidden in the boundaries. $\tan^{-1}(\sin(x))+\tan^{-1}(\csc(x))$ is a step function:
https://www.wolframalpha.com/input?i=arctan(sin(x))+++arctan(1/sin(x))=

Interesting...I suppose this makes it clear why you should simplify as much as possible to get to the meat of the integral. The graphs are only equal at the "peaks". Fascinating.

 

[SammyS]

Interesting...I suppose this makes it clear why you should simplify as much as possible to get to the meat of the integral. The graphs are only equal at the "peaks". Fascinating.

What peaks?

One issue here is that the constants of integration differ. Another issue is that the integrand for the integral
$\displaystyle \int \frac{\csc{x}\cot{x}}{1+\csc^2{x}}dx$
is undefined at $x = k\pi$, where $k$ is an integer.. On the other hand the integrand for
$\displaystyle \int \frac{\cos{x}}{1+\sin^2{x}}dx$
it is defined for all $x$ .

The graph suggested by @fresh_42 shows that
$\displaystyle \tan^{-1}(\sin(x))$
and
$\displaystyle -\tan^{-1}(\csc(x))$
differ by
$\pm \dfrac \pi 2$ .