When differentiating trigonometric function, we note the following:
$$\frac{d}{dx}(\sin x)=\cos x$$
$$\frac{d}{dx}(\cos x)=-\sin x$$
$$\frac{d}{dx}(\tan x)=\sec^2 x$$
$$\frac{d}{dx}(\sec x)=\sec x \tan x$$
$$\frac{d}{dx}(\csc x)=-\csc x \cot x$$
$$\frac{d}{dx}(\cot x)=-\csc^2 x$$
Since integration is the inverse of differentiation, it follows that
$$\int \cos x dx=\sin x+C$$
$$\int \sin x dx =-\cos x+C$$
$$\int \sec^2 x dx= \tan x+C$$
$$\int \csc x \cot x= -\csc x+C$$
$$\int \csc^2 =-\cot x+C$$
Example 1
Integrate $7\sec^2x$
Solution:
$\int 7\sec^2x=7\int\sec^2x$
$7(\tan x)+ C$
Example 2
Evaluate $\int \left(\cos x +7\sin x-\sec^2x\right) dx$
Solution:
$\sin x+ 7 (-\cos x)-\tan x + C$
$\sin x-7\cos x-\tan x + C$
Example 3
Evaluate $\int \left( 1+ \cos x +\csc^2 x\right) dx$
Solution:
Here, we use the power rule for 1 and we integrate the other using the normal procedure
$\frac{1x^{0+1}}{1}+\sin x +(-\cot x) +C$
$x+\sin x-\cot x +C$
Example 4
Integrate $1+2\cos x+9\csc x \cot x$
Solution:
$\int \left( 1+2\cos x+9\csc x \cot x\right) dx$
$1x+2\sin x +9 (-\csc x)+C$
$x+2\sin x -9\csc x+ C$
Example 5
Evaluate $\int \cos 3x dx$
Solution:
$\frac{1}{3}\sin 3x+C$
$\frac{\sin 3x}{3}+C$