Evaluate $\int \frac{1}{\sin x\cos x} dx $ Question: How to evaluate $\displaystyle \int \frac{1}{\sin x\cos x} dx $
I know that the correct answer can be obtained by doing:
$\displaystyle\frac{1}{\sin x\cos x} = \frac{\sin^2(x)}{\sin x\cos x}+\frac{\cos^2(x)}{\sin x\cos x} = \tan(x) + \cot(x)$ and integrating.
However, doing the following gets a completely different answer:
\begin{eqnarray*}
\int \frac{1}{\sin x\cos 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 x\cos 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?
 A: 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 x\left(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 x\left( 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 x\right)}{\cos x(1-\cos^2 x)}\\
& = & \frac{\sin x}{\cos x\>\sin^2 x} = \frac{1}{\cos x\sin x}.
\end{eqnarray*}
So I'm not sure why Mathematica says the second method is not "the right answer". 
A: This may be an easier method $$\int\frac{1}{\sin{x} \cdot \cos{x}} \ dx = \int\frac{\sec^{2}{x}}{\tan{x}} \ dx$$ by multiplying the numerator and denominator by $\sec^{2}{x}$
A: Taking log of $\rm\ sin^2(x)\ =\ 1 - cos^2(x)\ = (1-cos(x))\ (1+cos(x))\ $ shows both answers identical
A: 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}
A: Tangent half-angle substitution
$$\displaystyle \int \frac{1}{\sin x\cos x} dx=\displaystyle \int \frac{(1+t^2)}{t(1-t^2)} dt=\displaystyle \int \frac{1}{t} dt-\displaystyle \int \frac{1}{1-t} dt-\displaystyle \int \frac{1}{1+t} dt$$
A: $\sin(x)\cos(x) = \frac{1}{2} \sin(2x)$.
$I = 2\int \csc(2x)$ let $u = 2x$ then:
$I = \int \csc(u) du = - \log(\cot(2x) + \csc(2x)) + C$
A: Another way to integrate would be to let $\displaystyle t = \frac{\sin{x}}{\cos{x}}$, then $\displaystyle \frac{dt}{dx} = \frac{1}{\cos^2{x}} \Rightarrow dx = \cos^2{x}\;{dt}$. 
Thus $ \displaystyle I = \int\frac{\cos^2{x}}{\sin{x}}\;{dt} = \int\frac{\cos{x}}{\sin{x}}\;{dt} = \int\frac{1}{t}\;{dt} = \ln{t}+k = \ln\frac{\sin{x}}{\cos{x}}+k$. 
A: 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. 
A: Let $\int \frac{1}{\sin x \cos x}dx=\int \frac{2}{2 \sin x \cos x}dx$ 
=$\int \frac{2}{\sin 2x}dx$
Let $\tan x=t$
$\sin 2x=\frac{2t}{1+t^2}$
$\tan^{-1} t=x$
$\frac{1}{1+t^2}dt=dx$
$2\int \frac{dx}{\sin 2x} =2\int \frac{1+t^2}{2t}\cdot\frac{dt}{1+t^2}$
=$\int\frac{dt}{t}$
=$\ln t+C$
=$\ln (\tan x)+C$
A: $$\int \frac{1}{\sin x\cos x} dx  = \int( \tan x+\cot x)\, dx= \int\frac{\sin x}{\cos x}\, dx+\int\frac{\cos x}{\sin x} \,dx$$ from where it should be fairly simple....(choose correctly the variables though)
