What is the integral of 4cos(x)+9sin(x) ?

The integral of 4cos(x)+9sin(x) is 4sin(x)-9cos(x)+C

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integral of 4cos(x)+9sin(x)

\int 4 cos (x) + 9 sin (x) d x

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

Solution steps

\int 4 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 4 cos (x) d x + \int 9 sin (x) d x

\int 4 cos (x) d x = 4 sin (x)

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

= 4 sin (x) - 9 cos (x)

Add a constant to the solution

= 4 sin (x) - 9 cos (x) + C

\int \frac{5}{( 3 x + 5 ) ^{4}} d x

f (x) = \frac{4 x + 1}{2 x ^{2} + 1}

\int \frac{x - y}{x + y} d y

\int 9 x^{\frac{1}{2}} d x

\frac{d}{d x} (2 x^{2} + 6)

What is the integral of 4cos(x)+9sin(x) ?
The integral of 4cos(x)+9sin(x) is 4sin(x)-9cos(x)+C