What is the integral of (sin(x)-6cos(x)) ?

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

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integral of (sin(x)-6cos(x))

\int (sin (x) - 6 cos (x)) d x

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

Solution steps

\int sin (x) - 6 cos (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 sin (x) d x - \int 6 cos (x) d x

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

\int 6 cos (x) d x = 6 sin (x)

= - cos (x) - 6 sin (x)

Add a constant to the solution

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

\int_{1}^{\infty} \frac{2}{x} d x

\int \frac{y ^{2} + 1}{y} d y

\frac{d}{d x} ((x^{2} - 3 x + 4)^{4})

\frac{1}{1 - e ^{x}}

n = 0 \sum \infty \frac{3 + 2 n}{2 ^{n}}

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