# 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,

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$\cos\;2x=2\cos^2\;x-1$ – J. M. Dec 16 '10 at 15:03

## 3 Answers

HINT

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

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

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

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thanks for your answer.But , Can you write step - by -step solution ? – MAxcoder Dec 16 '10 at 15:09

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@MAx: Why not try replacing the $\cos\;2x$ first with one of AD's suggestions and see where it leads you? – J. M. Dec 16 '10 at 15:11

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@MAxcoder: I do not want to spoil the fun parts. – AD. Dec 16 '10 at 21:27
The answer for checking purposes is the following: $-\frac {\rm 1}{\rm 2}\rm{\cos(\rm 2x)}+\rm \ln(\cos(\rm x))$. – night owl Jul 5 '11 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

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

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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.

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