Answer
$H\left( x \right) = \tan x + \pi \sin x + C$
Work Step by Step
$$\eqalign{
& h\left( x \right) = {\sec ^2}x + \pi \cos x \cr
& {\text{Using the formulas in Table 2 }}\left( {{\text{see page 358}}} \right){\text{ }} \cr
& {\text{Function: }}cf\left( x \right) \to {\text{Particular antiderivative: }}cF\left( x \right) \cr
& {\text{Function: }}{\sec ^2}x \to {\text{Particular antiderivative: }}\tan x \cr
& {\text{Function: }}\cos x \to {\text{Particular antiderivative: }}\sin x \cr
& {\text{Function: }}f\left( x \right) + h\left( x \right) \to {\text{Particular antiderivative: }}f\left( x \right) + h\left( x \right) \cr
& {\text{And applying the Theorem 1}},{\text{ we obtain}} \cr
& h\left( x \right) = {\sec ^2}x + \pi \cos x \cr
& H\left( x \right) = \tan x + \pi \sin x + C \cr
& \cr
& {\text{Checking the answer by differentiation}} \cr
& H'\left( x \right) = \frac{d}{{dx}}\left[ {\tan x} \right] + \pi \frac{d}{{dx}}\left[ {\sin x} \right] + \frac{d}{{dx}}\left[ C \right] \cr
& H'\left( x \right) = {\sec ^2}x + \pi \left( {\cos x} \right) + 0 \cr
& H'\left( x \right) = {\sec ^2}x + \pi \cos x \cr} $$
