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$$\int\sec^2(4x)\tan^2(4x)\,\mathrm{d}x$$

This is the original formula.

I used a U-substitution $u=4x$ so that means $ \frac{\mathrm{d}u}4=\mathrm{d}x $

So assuming I'm right then...

$$\frac14\int\sec^2(u)\tan^2(u)\,\mathrm{d}u$$

So I thought this would mean that after you take the integral you would have

$$-\frac14\frac{\tan^3(4x)\ln^3|4x|}{36}+C$$

However my webwork is telling me my answer is dead wrong. I believe it is because I messed up the product rule... but then I checked on a website but it wouldn't explain its answer with out money. So could some one work out the problem so I can see the proper integral? I am having trouble understanding how I could reverse the product rule. Do I have to use another substitution? Could I do this without substituting the trig function?

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  • $\begingroup$ Could you show your work about getting your result? $\endgroup$ – xbh Aug 11 '18 at 5:01
  • $\begingroup$ I honestly don't know how to show it any further than I put it in the post. I put exactly what I have written on my paper in this post. Even in the same order exactly. $\endgroup$ – Disgruntled Student Aug 11 '18 at 5:05
  • $\begingroup$ Yeah, but we want know that how did you get your last line from $\int \sec^2(u)\tan^2(u)\mathrm du$? You post it then we can help you to point out where you were wrong. $\endgroup$ – xbh Aug 11 '18 at 5:08
  • $\begingroup$ I see. I did notice I did not do the product rule. I think I might be kinda confused on how to reverse the product rule. $\endgroup$ – Disgruntled Student Aug 11 '18 at 5:22
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$$\frac14\int\sec^2(u)\tan^2(u)\,\mathrm{d}u$$ put $$\tan(u)=t$$ thus $$\sec^2(u)\,\mathrm{d}u=\mathrm{d}t$$ $$\frac14\int t^2\,\mathrm{d}t$$

$$\frac{t^3}{12}+C$$ $$\frac{\tan^3(u)}{12}+C$$ $$\frac{\tan^3(4x)}{12}+C$$

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  • $\begingroup$ All I do during the second substitution is to try to manipulate the trig function to be the same as the other trig function in the integral? $\endgroup$ – Disgruntled Student Aug 11 '18 at 5:09
  • $\begingroup$ @DisgruntledStudent You have to make sure $du$ matches as well. $\endgroup$ – Toby Mak Aug 11 '18 at 7:43
  • $\begingroup$ how do you not end up with a chain-rule effect with u? wouldn't $dt=sec^2(u)(1)du $ so wouldn't that ruin taking the derivative ? $\endgroup$ – Disgruntled Student Aug 12 '18 at 4:38
  • $\begingroup$ @DisgruntledStudent After you substitute, you have to make sure it converts to an integral you know. Polynomials, trigonometric functions like $\sin u$, $\cos u$, $\tan u$ or their inverses/reciprocals are all examples. $\endgroup$ – Toby Mak Aug 14 '18 at 8:16
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Quick method:

$$ \int \tan^2(x)\sec^2(x) \mathrm dx = \int \tan^2(x) \mathrm d(\tan (x)) = \cdots $$

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Hint: $$\frac{1}{4} \int \sec^2(u)\tan^2(u)\ du=\frac{1}{4} \int (1+\tan^2(u))\tan^2(u)\ du$$ Now let $\tan u=w$.

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  • $\begingroup$ Is this my only option? Must I substitute the trig value? $\endgroup$ – Disgruntled Student Aug 11 '18 at 5:06
  • $\begingroup$ It's the simplest! $\endgroup$ – Nosrati Aug 11 '18 at 5:07

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