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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

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 (r)

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

Simplify

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

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

Add a constant to the solution

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

\int x^{2} sin (4 x^{3} + 7) d x

f (x) = 18 x - 3 x^{2}

\int \frac{1}{z ^{2} 9 - z ^{2}} d z

\frac{4 s + 2}{s ^{2} + 8 s + 16}

\int \frac{1}{x ^{5} + 2 x ^{3} + x} d x

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