What is the integral of (7-1/3 sin(x)+9cos(x)) ?

The integral of (7-1/3 sin(x)+9cos(x)) is 7x+1/3 cos(x)+9sin(x)+C

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integral of (7-1/3 sin(x)+9cos(x))

\int (7 - \frac{1}{3} sin (x) + 9 cos (x)) d x

7 x + \frac{1}{3} cos (x) + 9 sin (x) + C

Solution steps

\int 7 - \frac{1}{3} sin (x) + 9 cos (x) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 7 d x - \int \frac{1}{3} sin (x) d x + \int 9 cos (x) d x

\int 7 d x = 7 x

\int \frac{1}{3} sin (x) d x = - \frac{1}{3} cos (x)

\int 9 cos (x) d x = 9 sin (x)

= 7 x - (- \frac{1}{3} cos (x)) + 9 sin (x)

Simplify

= 7 x + \frac{1}{3} cos (x) + 9 sin (x)

Add a constant to the solution

= 7 x + \frac{1}{3} cos (x) + 9 sin (x) + C

f (x) = x + \frac{17}{x}, at x = 1

\frac{d}{d x} (\frac{x}{x + 19})

\frac{d y}{d x} = - \frac{4 sin ( x ) cos ( x )}{y}

x \to 2 lim (x^{2} - 2 x + 1)

\int ln (x) \cdot e^{- 2 x} d x

What is the integral of (7-1/3 sin(x)+9cos(x)) ?
The integral of (7-1/3 sin(x)+9cos(x)) is 7x+1/3 cos(x)+9sin(x)+C