1
$\begingroup$

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

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?

|cite|improve this question
$\endgroup$
  • $\begingroup$ 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
5
$\begingroup$

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

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

|cite|improve this answer
$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

Quick method:

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

|cite|improve this answer
$\endgroup$
1
$\begingroup$

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 $\tan u=w$.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.