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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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4 Answers 4

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

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

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

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  • $\begingroup$ Please format your posts using MathJax to make them easier to read. $\endgroup$
    – naslundx
    Apr 15, 2014 at 8:39

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