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?
Thanks.
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.