I had this question: $$\int 3x\sec^2(4x) dx$$

After doing integration by parts for the first time, setting $u=3x$ and $dv=\sec^2(4x)dx$ and doing derivative and integral, I got $$\int 3x\sec^2(4x) dx = \frac{3}{4}x\tan(4x)-\frac{3}{4}\int \tan(4x) dx$$

At this point, I realize I can solve for the $\tan(4x)$ using $u$-substitution, but I continue doing integration by parts.

For $$\int \tan(4x) dx$$ I substitute $u=\tan(4x)$ and $dv=1dx$ getting $$\int \tan(4x) dx = \tan(4x)\cdot x - \int x\cdot 4\sec^2(4x)dx$$

Substituting everything back in, I get $$\int 3x\sec^2(4x) dx = \frac{3}{4}x\tan(4x) - \frac{3}{4}\left(\tan(4x)\cdot x -\int x\cdot 4\sec^2(4x)dx\right)$$

Distributing the $3/4$ and simplifying, I get $0=0$ — that's not the answer. I don't believe I broke any rules using integration by parts, yet the answer is invalid. Why doesn't this work?


  • 8
    $\begingroup$ I'd just like to point out that, in fact, zero does equal zero. $\endgroup$ – law-of-fives Feb 6 '18 at 4:39
  • $\begingroup$ :/............... $\endgroup$ – Art Feb 6 '18 at 4:40
  • $\begingroup$ I didn't read your post in detail, but I would bet on a sign error $\endgroup$ – qbert Feb 6 '18 at 4:43
  • 6
    $\begingroup$ But Art, I hope you realise that law-of-fives was making a serious point. You have not got an invalid answer. You have not made any algebraic mistakes (I assume - didn't actually check), and the answer you have got is a correct equation. It happens to be useless because it doesn't tell you the value of the integral you want, but that's a different thing... $\endgroup$ – David Feb 6 '18 at 4:45
  • 5
    $\begingroup$ Your second integration by parts is not wrong, but it's taking you backwards: first you had $dv = \sec^24x dx$ to get $v = \frac14 \tan 4x$, and then you set $u = \tan 4x$ to get $du = 4 \sec^24x dx$. This is like multiplying by 2 then dividing by 2. $\endgroup$ – Théophile Feb 6 '18 at 4:47

You proved that your integral is equal to itself. This is a true statement! It's not a useful thing to do, but the integration by parts formula doesn't come with a guarantee that it will produce something useful.

In general, if we apply integration by parts twice, we could always get back to where we started:

$$ \int u \, dv = uv - \int v \, du = uv - \left(uv - \int u \,dv\right)=\int u \, dv $$

This is essentially what you just did. The $u$ in your second integration by parts is equal to the $v$ in your first one, and vice versa.

  • $\begingroup$ Makes sense. So, I would have to do u-sub for tan(4x)? $\endgroup$ – Art Feb 6 '18 at 4:47
  • $\begingroup$ That seems like a good way to proceed, yes. $\endgroup$ – Micah Feb 6 '18 at 4:48

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.