Question:

<div>$\int \cos x dx = \sin x + C \Rightarrow \int \cos(5+4x) dx = \frac{\sin(5+4x)}{4} + C$<br>Similarly $\int \sec^2(3-4x) dx = \frac{\tan(3-4x)}{-4} + C$<br>$\therefore \int [\cos(5+4x) + \sec^2(3-4x)] dx = \frac{\sin(5+4x)}{4} - \frac{\tan(3-4x)}{4} + C$</div>