What is the integral of (8cos(x)+9sin(x)) ?

The integral of (8cos(x)+9sin(x)) is 8sin(x)-9cos(x)+C

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integral of (8cos(x)+9sin(x))

\int (8 cos (x) + 9 sin (x)) d x

8 sin (x) - 9 cos (x) + C

Solution steps

\int 8 cos (x) + 9 sin (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 8 cos (x) d x + \int 9 sin (x) d x

\int 8 cos (x) d x = 8 sin (x)

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

= 8 sin (x) - 9 cos (x)

Add a constant to the solution

= 8 sin (x) - 9 cos (x) + C

\int - \frac{1}{4} csc (8 t) d t

\frac{\partial}{\partial x} (x^{2} + y e^{x})

\int \frac{11 - 2 x}{x ^{2} + 9 x + 14} d x

\frac{2 x}{3 - cot ( x )}

x \to 3 lim ((x^{2} - 9)^{x - 3})

What is the integral of (8cos(x)+9sin(x)) ?
The integral of (8cos(x)+9sin(x)) is 8sin(x)-9cos(x)+C