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Question: How to evaluate $displaystyle int frac{1}{sin xcos x} dx $

I know that the correct answer can be obtained by doing:

$displaystylefrac{1}{sin xcos x} = frac{sin^2(x)}{sin xcos x}+frac{cos^2(x)}{sin xcos x} = tan(x) + cot(x)$ and integrating.

However, doing the following gets a completely different answer:
begin{eqnarray*}
int frac{1}{sin xcos x} dx

&=&int frac{sin x}{sin^2(x)cos x} dx\
&=&int frac{sin x}{(1-cos^2(x))cos x} dx.
end{eqnarray*}
let $u=cos x, du=-sin x dx$; then
begin{eqnarray*}
int frac{1}{sin xcos x} dx

&=&int frac{-1}{(1-u^2)u} du\

&=&int frac{-1}{(1+u)(1-u)u}du\
&=&int left(frac{-1}{u} - frac{1}{2(1-u)} + frac{1}{2(1+u)}right) du\
&=&-ln|cos x|+frac{1}{2}ln|1-cos x|+frac{1}{2}ln|1+cos x|+C
end{eqnarray*}

I tested both results in Mathematica, and the first method gets the correct answer, but the second method doesn't. Is there any reason why this second method doesn't work?

If I take the derivative of your second answer (call it $g(x)$), I get:
begin{eqnarray*}
frac{dg}{dx}
& = & -frac{-sin x}{cos x} + frac{sin x}{2(1-cos x)} + frac{-sin x}{2(1+cos x)}\
& = & frac{sin xleft(1-cos^2 x + frac{1}{2}cos x(1+cos x) - frac{1}{2}cos x(1-cos x)right)}{cos x(1-cos x)(1+cos x)}\
& = & frac{sin xleft( 1- cos^2 x + frac{1}{2}cos x + frac{1}{2}cos^2 x - frac{1}{2}cos x + frac{1}{2}cos^2 xright)}{cos x(1-cos^2 x)}\
& = & frac{sin x}{cos x>sin^2 x} = frac{1}{cos xsin x}.
end{eqnarray*}
So I'm not sure why Mathematica says the second method is not "the right answer".

This may be an easier method $$intfrac{1}{sin{x} cdot cos{x}} dx = intfrac{sec^{2}{x}}{tan{x}} dx$$ by multiplying the numerator and denominator by $sec^{2}{x}$

Taking log of $rm sin^2(x) = 1 - cos^2(x) = (1-cos(x)) (1+cos(x)) $ shows both answers identical

The second method gives the same answer as the first. By the first method, the answer you get is $-log(cos x) + log(sin x)$. The first term is the same as what you get by the second method.

What you need to show is that $log(sin x) = frac{1}{2}log(1-cos x) + frac{1}{2}log(1+cos x)$.

begin{equation}
begin{split}
frac{1}{2}log(1-cos x) + frac{1}{2}log(1+cos x) &= frac{1}{2}left( log(2 sin^2 frac{x}{2}) + log(2 cos^2 frac{x}{2})right)\ & = log(2 sin frac{x}{2} cos frac{x}{2})
end{split}
end{equation}

Tangent half-angle substitution

$$displaystyle int frac{1}{sin xcos x} dx=displaystyle int frac{(1+t^2)}{t(1-t^2)} dt=displaystyle int frac{1}{t} dt-displaystyle int frac{1}{1-t} dt-displaystyle int frac{1}{1+t} dt$$

$sin(x)cos(x) = frac{1}{2} sin(2x)$.

$I = 2int csc(2x)$ let $u = 2x$ then:

$I = int csc(u) du = - log(cot(2x) + csc(2x)) + C$

Integrand $ =dfrac {1}{sin(x)cos(x)} = 2 csc 2x $

Its integral is obtained by direct application of listed standard trigonometric function integration formulae. Using chain rule for constant double angle:

$$ 2 log (tan dfrac{2 x}{2}) cdot frac12= log (tan x ) + c $$

agrees with OP's second result when it is further simplified.