What is the integral of 2cos(x)-9sin(x) ?

The integral of 2cos(x)-9sin(x) is 2sin(x)+9cos(x)+C

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integral of 2cos(x)-9sin(x)

\int 2 cos (x) - 9 sin (x) d x

2 sin (x) + 9 cos (x) + C

Solution steps

\int 2 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 2 cos (x) d x - \int 9 sin (x) d x

\int 2 cos (x) d x = 2 sin (x)

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

= 2 sin (x) - (- 9 cos (x))

Simplify

= 2 sin (x) + 9 cos (x)

Add a constant to the solution

= 2 sin (x) + 9 cos (x) + C

\int \frac{4}{x 16 - ln ( x ^{2} )} d x

\int 3 x^{\frac{4}{3}} d x

\int x + \frac{2}{x} d x

\int \frac{3 x ^{4} - 9 x ^{2} + 1}{x ^{2}} d x

\int 3 e^{x} + 4 d x

What is the integral of 2cos(x)-9sin(x) ?
The integral of 2cos(x)-9sin(x) is 2sin(x)+9cos(x)+C