# How can I find $\int\tan\;x\;\cos\;2x\;\mathrm dx$?

My question is ; How can I solve the following integral question?

$$\int\tan\;x\;\cos\;2x\;\mathrm dx$$

• $\cos\;2x=2\cos^2\;x-1$ Dec 16, 2010 at 15:03

HINT

(1) $\cos 2x = \cos^2x-\sin^2x=2\cos^2x-1$.

(2) $2\sin x\cos x = \sin 2x$.

(3) $\frac{d}{dx}\log(f(x))=?$

• thanks for your answer.But , Can you write step - by -step solution ? Dec 16, 2010 at 15:09
• @MAx: Why not try replacing the $\cos\;2x$ first with one of AD's suggestions and see where it leads you? Dec 16, 2010 at 15:11
• @MAxcoder: I do not want to spoil the fun parts. Dec 16, 2010 at 21:27
• The answer for checking purposes is the following: $-\frac {\rm 1}{\rm 2}\rm{\cos(\rm 2x)}+\rm \ln(\cos(\rm x))$. Jul 5, 2011 at 12:03

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

$\displaystyle \int \cot x \ \ (1 + \sin^5 x) \ \ \text{dx}$ ?

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.

$\int\tan\;x\;\cos\;2x\;\mathrm dx$

=$\int\frac{sinx}{cosx}\;\cos\;2x\;\mathrm dx$

=$-\int\frac{1}{cosx}\;\cos\;2x\;\mathrm -sinx dx$ $\frac{d(cosx)}{dx}$ = -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]

• Please format your posts using MathJax to make them easier to read. Apr 15, 2014 at 8:39