Why does this have to be integrated by parts? I'm asked to perform the following integration using integration by parts: 
$$\int \frac{\sin(2t+3)}{\sqrt{1-\sin^2(2t+3)}}$$ 
However, when I look at the denominator I see an identity, and my instinct is just to do: 
$$\frac{\sin(2t+3)}{\sqrt{1-\sin^2(2t+3)}} = \frac{\sin(2t+3)}{\cos(2t+3)} = \tan(2t+3)$$ 
and then: 
$\int \tan(2t+3) = -\frac{1}{2}\ln|\cos(2t+3)| + C$
However, the answer given is $$\frac{1}{2\cos(2t+3)} = \frac{1}{2} \frac{1}{\cos(2t+3)}$$
Clearly, my aforementioned instinct has led me astray here - but why? 
 A: Your answer is also incorrect given the problem.
$$\int \! \frac{\sin(2t+3)}{\sqrt{1-\sin^2(2t+3)}}\mathrm{d}t$$ 
You may make the following simplification,
$$\frac{\sin(2t+3)}{\sqrt{1-\sin^2(2t+3)}} = \frac{\sin(2t+3)}{|\cos(2t+3)|}$$ 
but not what you originally had written. For certain values of $t$ your integral would not agree with your result. 
Focusing on:
$$\int \!  \frac{\sin(2t+3)}{|\cos(2t+3)|}\mathrm{d}t$$ 
we can consider two cases. One is when $\cos(2t+3) > 0$ and another when $\cos(2t+3) < 0.$ We can safely assume that we aren't going to integrate over any singularities.
There is a simple fix to your result. Naturally, the sign of $\cos(2t+3)$ is going to stay the same throughout the integral so our answer is going to be either
$$\int \!  \frac{\sin(2t+3)}{\cos(2t+3)}\mathrm{d}t \;\; \text{or} \;\;\int \!  -\frac{\sin(2t+3)}{\cos(2t+3)}\mathrm{d}t.$$ 
If $\cos(2t+3)$ is positive on the interval of integration our answer will be $-\frac{1}{2}\log(\cos(2t+3))$ and if it is negative on the interval our answer will be $\frac{1}{2}\log(-\cos(2t+3)).$ Both of the discrepancies can be resolved with the sign function or $\operatorname{sgn}(x)$. With this function (which will be explained), we can write
$$\int \! \frac{\sin(2t+3)}{\sqrt{1-\sin^2(2t+3)}}\mathrm{d}t = -\dfrac{1}{2}\operatorname{sgn}(\cos(2t+3))\log(|\cos(2t+3)|) + C.$$ 
This answer is almost certainly unnecessary in your case but it is important to consider when our answers do and do not make sense and how to make our answers applicable to more cases.
The sign function simply returns $1$ if the input is positive, $-1$ if it is negative and (usually) $0$ if the input is $0$. In your case you could replace $\operatorname{sgn}(\cos(2t+3))$ with $\dfrac{\cos(2t+3)}{|\cos(2t+3)|}$ but this is generally messier. You cannot do this in general because the denominator could still be zero but in this case it makes sense.
Here is a graph of $\operatorname{sgn}(x)$ (from wikipedia):

