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.

2. Originally Posted by Dranalion
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.
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.

3. Originally Posted by Dranalion
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.
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?

4. The above solution is cleaner than what i posted.

5. 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.

6. 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.