What is the integral of cos^{18}(x)sin(x) ?

The integral of cos^{18}(x)sin(x) is -(cos^{19}(x))/(19)+C

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integral of cos^{18}(x)sin(x)

\int cos^{18} (x) sin (x) d x

- \frac{cos ^{19} ( x )}{19} + C

Solution steps

\int cos^{18} (x) sin (x) d x

Apply u-substitution

: \int - u^{18} d u

= \int - u^{18} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int u^{18} d u

Apply the Power Rule

: \frac{u ^{19}}{19}

= - \frac{u ^{19}}{19}

Substitute back

u = cos (x)

= - \frac{cos ^{19} ( x )}{19}

Add a constant to the solution

= - \frac{cos ^{19} ( x )}{19} + C

- 5 (7 x^{2} + 4)^{- 6}

z \to 10 lim (\frac{1}{z - 10})

\int arccos (\frac{1}{x}) d x

y = 8 x^{2} - 2 x, y = x^{3} - 8 x + 3

\frac{d y}{d x} = \frac{y ^{2} + x y + x ^{2}}{x ^{2}}

What is the integral of cos^{18}(x)sin(x) ?
The integral of cos^{18}(x)sin(x) is -(cos^{19}(x))/(19)+C