What is the integral of sin(x)-8cos(x) ?

The integral of sin(x)-8cos(x) is -cos(x)-8sin(x)+C

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integral of sin(x)-8cos(x)

\int sin (x) - 8 cos (x) d x

- cos (x) - 8 sin (x) + C

Solution steps

\int sin (x) - 8 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 8 cos (x) d x

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

\int 8 cos (x) d x = 8 sin (x)

= - cos (x) - 8 sin (x)

Add a constant to the solution

= - cos (x) - 8 sin (x) + C

f (x) = cot (7 x)

x \to 0 lim \frac{( \frac{1}{( x + 4 )} ) - ( \frac{1}{4} )}{x}

5 x + 8

y^{^{'}} = \frac{x}{1 + x ^{2}}

\int 64 sin^{3} (x) d x

What is the integral of sin(x)-8cos(x) ?
The integral of sin(x)-8cos(x) is -cos(x)-8sin(x)+C