How can I find $\int\tan\;x\;\cos\;2x\;\mathrm dx$? My question is ; How can I solve the following integral question?
$$\int\tan\;x\;\cos\;2x\;\mathrm dx$$
Thanks in advance,
 A: HINT
(1) $\cos 2x = \cos^2x-\sin^2x=2\cos^2x-1$.
(2) $2\sin x\cos x = \sin 2x$.
(3) $\frac{d}{dx}\log(f(x))=?$
A: Suppose I gave you an integral of the form
$\displaystyle \int \cot x \ \ f(\sin x) \ \text{dx}$
Can you think of a substitution to get rid of the $\cot x$ term?
For a concrete example, can you try evaluating
$\displaystyle \int \cot x  \ \ (1 + \sin^5 x) \ \ \text{dx}$ ?
A: I'm going to tell you that by parts done directly isn't the way to approach this:
$$\int \tan(x)\cos(2x)dx = -\ln(\cos(x))\cos(2x) - 2\int \ln(\cos(x))\sin(2x)dx$$
As you can see, this expression is not likely to become any more manageable by solving the next integral.
In short, your problem comes down to simplifying the expression $\tan(x)cos(2x)$. Big hint. The other answers have shown you how to do this. Once you simplify it, you will have a much easier job of integrating said expression and you most certainly won't need integration by parts.
A: $\int\tan\;x\;\cos\;2x\;\mathrm dx$
=$\int\frac{sinx}{cosx}\;\cos\;2x\;\mathrm dx$
=$-\int\frac{1}{cosx}\;\cos\;2x\;\mathrm -sinx dx$                    $\frac{d(cosx)}{dx}$ = -sinx.dx
=$-\int\frac{1}{cosx}\;\cos\;2x\;\mathrm d(cosx)$                               
cos2x=2${cos^2x}$-1
=$-\int\frac{1}{cosx}\cdot(2{cos^2x}$-1) d(cosx)$
=$-\int(2{cosx}-\frac{1}{cosx}) d(cosx)$
= $-[\int(2{cosx}d(cosx)$ - $\int\frac{d(cosx)}{cosx}$
= $-[{cos^2x} + C1 - log (cosx) -C2]$
=$log(cosx)-{cos^2x}+C2-C1$ [C2,C1 - Integral Constants]
