Integrating $\sec^{2}(x)(\ln(\tan(x)+2))dx$

Question

I'm going through some Calculus homework with using integration by parts and I'm getting hit with this sucker:

$$\int\sec^2(x)\ln(\tan(x)+2)dx$$

I'm not sure how to go about tackling this monster. Things I've tried:

1)Letting $u$ equal the entire function, and $dv$ equal $dx$; problem I ran into here is I then feel uncomfortable obtaining the derivative of $u$

2) Trying to do u-sub first before doing parts, but again with the uncomfortability/doubting my steps

That's honestly about it (wish I could give more feedback), first thing I did try though was to let $u$ equal everything inside $sec^2$ but realized that was illegal. So, any suggestions to kick start me on the set up of this beast? I don't necessarily need it solved (though feel free to), but I'm mainly looking at the set up, I just don't know where to begin.

Answer 1

Make a substitution first \begin{eqnarray*} \int \sec^2(x) \ln( \tan(x)+2) dx = ? \end{eqnarray*} Let $u= \tan(x)+2$ so $du= \sec^2(x) dx$ & the integral becomes \begin{eqnarray*} \int \ln( u) du \end{eqnarray*} Now use integration by parts ... integrate 1 & differentiate $ \ln $ \begin{eqnarray*} \int \ln( u) du = u \ln u -u +c \end{eqnarray*} Now substitute back & we have \begin{eqnarray*} \int \sec^2(x) \ln( \tan(x)+2) dx =( \tan(x)+2) \ln( \tan(x)+2)-( \tan(x)+2)+c \end{eqnarray*}

Answer 2

My guess is that the logarithm is not inside the secant. I don't think the function can be integrated in any simple closed form if that's the case; as such, I am going to assume your worksheet meant
$$\int \sec^2(x)\log(\tan(x)+2)dx$$ We know the integral of $\sec^2(x)$ and we know the derivative of $\log(x)$, so let's use those. Let $v' = \sec^2(x) \implies v = \tan(x)$ and let $u = \log(\tan(x)+2) \implies u' = \frac{\sec^2(x)}{\tan(x)+2}$
This makes our integral $$=\log(\tan(x)+1)\tan(x)-\int \frac{\tan(x)\sec^2(x)}{\tan(x)+2}$$ Now we let $u=\tan(x)$ and $du=\sec^2(x)dx$ $$=\log(\tan(x)+1)\tan(x)-\int \frac{u}{u+2}du$$ Now we do partial fraction decomposition, noting that $\frac{u}{u+2} = 1-\frac{2}{u+2}$ to get two integrals we can manage. Back substitute for $u$ and you are done