What is the integral of 8sin^3(t)cos(t) ?

The integral of 8sin^3(t)cos(t) is 2sin^4(t)+C

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integral of 8sin^3(t)cos(t)

\int 8 sin^{3} (t) cos (t) d t

2 sin^{4} (t) + C

Solution steps

\int 8 sin^{3} (t) cos (t) d t

Take the constant out:

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

= 8 \cdot \int sin^{3} (t) cos (t) d t

Apply u-substitution

: \int u^{3} d u

= 8 \cdot \int u^{3} d u

Apply the Power Rule

: \frac{u ^{4}}{4}

= 8 \cdot \frac{u ^{4}}{4}

Substitute back

u = sin (t)

= 8 \cdot \frac{sin ^{4} ( t )}{4}

Simplify

8 \cdot \frac{sin ^{4} ( t )}{4} : 2 sin^{4} (t)

= 2 sin^{4} (t)

Add a constant to the solution

= 2 sin^{4} (t) + C

y = cos^{2} (t)

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

f (x) = (3 x^{3})^{0.5}

\frac{\partial}{\partial y} (x y^{3} - x^{2} y)

2 x y^{^{'}} = cos^{2} (y)

What is the integral of 8sin^3(t)cos(t) ?
The integral of 8sin^3(t)cos(t) is 2sin^4(t)+C