Question

find the indefinite integral and check the result by differentiation. (remember the constant of integration.)
int(3 cos (x)+8 sin (x)) d x

Explanation:

Step1: Apply the sum rule of integration

The sum rule states that \(\int(f(x)+g(x))dx=\int f(x)dx+\int g(x)dx\). So, \(\int(3\cos(x)+8\sin(x))dx = 3\int\cos(x)dx+8\int\sin(x)dx\).

Step2: Integrate each term

We know that \(\int\cos(x)dx=\sin(x)+C_1\) and \(\int\sin(x)dx=-\cos(x)+C_2\).
So, \(3\int\cos(x)dx+8\int\sin(x)dx=3\sin(x)-8\cos(x)+C\) (where \(C = 3C_1+8C_2\) is the constant of integration).

Step3: Check by differentiation

Differentiate \(y = 3\sin(x)-8\cos(x)+C\) using the sum rule \((u + v+w)'=u'+v'+w'\) and the basic differentiation rules \((\sin(x))'=\cos(x)\) and \((\cos(x))'=-\sin(x)\).
\(y'=(3\sin(x))'-(8\cos(x))'+C'\)
\(y'=3\cos(x)-8(-\sin(x))+0\)
\(y'=3\cos(x)+8\sin(x)\)

Answer:

\(3\sin(x)-8\cos(x)+C\)