Ta có $I=\int\frac{\sin \alpha\cos x+\cos \alpha\sin x}{\cos^2 x}dx=\int\limits \frac{\sin \alpha dx}{\cos x}+\int\limits \frac{\cos \alpha sin xdx}{\cos^2 x} $
$I=\int\frac{\sin\alpha
dx}{\sin(x+\frac{\pi}{2})}-\int\frac{\cos\alpha d\cos x}{\cos^2 x}$
$=\int\frac{\sin\alpha
dx}{2\tan(\frac{x}{2}+\frac{\pi}{4})\cos^2(\frac{x}{2}+\frac{\pi}{4})}+\frac{\cos\alpha}{\cos
x}$
$=\int\frac{\sin\alpha
d\tan(\frac{x}{2}+\frac{\pi}{4})}{\tan(\frac{x}{2}+\frac{\pi}{4})}+\frac{\cos\alpha}{\cos
x}$
$I=\sin \alpha.\ln|\tan (\frac{x}{2}+ \frac{\pi}{4} )|+\frac{\cos \alpha}{\cos x}+C(ycbt) $