Hello,

I am stuck with this particular integral:

∫ tan^4(2x)sec^4(2x)dx

As far as I could go was to:

∫ (tan^2(2x)sec^2(2x))^2 dx

Any help is much appreciated,

Dranalion.

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- May 8th 2009, 04:37 PMDranalionAntiderivative help, please.
Hello,

I am stuck with this particular integral:

∫ tan^4(2x)sec^4(2x)dx

As far as I could go was to:

∫ (tan^2(2x)sec^2(2x))^2 dx

Any help is much appreciated,

Dranalion. - May 8th 2009, 05:02 PMChris L T521
Hint:

$\displaystyle \begin{aligned}\int\tan^4\!\left(2x\right)\sec^4\! \left(2x\right)\,dx & = \int\tan^4\!\left(2x\right)\sec^2\!\left(2x\right) \sec^2\!\left(2x\right)\,dx\\ & = \int\tan^4\!\left(2x\right)\left(\tan^2\!\left(2x\ right)+1\right)\sec^2\!\left(2x\right)\,dx\\ & = \int\left(\tan^6\!\left(2x\right)+\tan^4\!\left(2x \right)\right)\sec^2\!\left(2x\right)\,dx\end{alig ned}$

Now make the substitution $\displaystyle u=\tan\!\left(2x\right)$

Can you continue? - May 8th 2009, 05:02 PMBanned for attempted hacking
Suggestion

$\displaystyle \int (tan^{2}(2x)sec^2(2x))^{2} dx$=

$\displaystyle \int (tan^{2}(2x) ( 1+tan^{2}(2x) )^{2} dx$

$\displaystyle \int (tan^{4}(2x)+tan^{2}(2x) )^{2} dx$

Multiply out

$\displaystyle \int (tan^{8}(2x)+2tan^{6}(2x) + tan^{4}(2x)) dx$

Use the reduction formula for each of the powers of tanx.

There might be a cleaner way to do this though. - May 8th 2009, 05:03 PMBanned for attempted hacking
The above solution is cleaner than what i posted.

- May 9th 2009, 02:17 AMDranalion
I understand the process you have used to find an easier integral to evaluate, and also, from the textbook, I know the answer is going to be:

[5*tan^7(2x) + 7*tan^5(2x)] / 70

However, I do not understand the method used to get there!

Can anyone please assist me here?!

Any help much appreciated, again,

Dranalion. - May 9th 2009, 11:36 AMBanned for attempted hacking
You have to remember that the answer in your book is simplified.

If you actually evaluate the integral you should get a similar answer which you could simplify; and they would end up been exactly equivalent.

If you don't get the same answer it's alright as long as the logic is correct and there are no mistakes.