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

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

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

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

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

Solution steps

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

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

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

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

Simplify

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

Add a constant to the solution

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

\int (9 x^{2} - 9 x + 2) d x

\int \frac{3 x ^{4} - 4 x + 7}{2 x ^{2}} d x

\int \frac{sin ( x )}{( 5 - cos ( x ) ) ^{3}} d x

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

\int 8 (5^{x}) d x

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