What is the integral of 8sin^4(x)cos(x) ?

The integral of 8sin^4(x)cos(x) is 8/5 sin^5(x)+C

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integral of 8sin^4(x)cos(x)

\int 8 sin^{4} (x) cos (x) d x

\frac{8}{5} sin^{5} (x) + C

Solution steps

\int 8 sin^{4} (x) cos (x) d x

Take the constant out:

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

= 8 \cdot \int sin^{4} (x) cos (x) d x

Apply u-substitution

: \int u^{4} d u

= 8 \cdot \int u^{4} d u

Apply the Power Rule

: \frac{u ^{5}}{5}

= 8 \cdot \frac{u ^{5}}{5}

Substitute back

u = sin (x)

= 8 \cdot \frac{sin ^{5} ( x )}{5}

Simplify

8 \cdot \frac{sin ^{5} ( x )}{5} : \frac{8}{5} sin^{5} (x)

= \frac{8}{5} sin^{5} (x)

Add a constant to the solution

= \frac{8}{5} sin^{5} (x) + C

\frac{d}{d x} (\frac{2}{2 - x})

\int \frac{( x ^{2} )}{x ^{2} + 1} d x

(y^{4} x) \frac{d y}{d x} = 1 + x, y (1) = 2

\frac{d}{d x} (5 - 2 (x - 2)^{2})

\frac{d}{d y} (36 - y^{2})

What is the integral of 8sin^4(x)cos(x) ?
The integral of 8sin^4(x)cos(x) is 8/5 sin^5(x)+C