What is the integral of (2cos(x)+4sin(x)+6) ?

The integral of (2cos(x)+4sin(x)+6) is 2sin(x)-4cos(x)+6x+C

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integral of (2cos(x)+4sin(x)+6)

\int (2 cos (x) + 4 sin (x) + 6) d x

2 sin (x) - 4 cos (x) + 6 x + C

Solution steps

\int 2 cos (x) + 4 sin (x) + 6 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 4 sin (x) d x + \int 6 d x

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

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

\int 6 d x = 6 x

= 2 sin (x) - 4 cos (x) + 6 x

Add a constant to the solution

= 2 sin (x) - 4 cos (x) + 6 x + C

\int_{1}^{2} \frac{x ^{3} + 3 x ^{6}}{x ^{4}} d x

\int_{0}^{2} 5 x d x

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

x \to - \infty lim (a + \frac{1}{x})

(7 x^{2} + 6 x) (7 x - 8)

What is the integral of (2cos(x)+4sin(x)+6) ?
The integral of (2cos(x)+4sin(x)+6) is 2sin(x)-4cos(x)+6x+C