What is the integral of-17sin^4(t)cos(t) ?

The integral of-17sin^4(t)cos(t) is -17/5 sin^5(t)+C

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integral of-17sin^4(t)cos(t)

\int - 17 sin^{4} (t) cos (t) d t

- \frac{17}{5} sin^{5} (t) + C

Solution steps

\int - 17 sin^{4} (t) cos (t) d t

Take the constant out:

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

= - 17 \cdot \int sin^{4} (t) cos (t) d t

Apply u-substitution

: \int u^{4} d u

= - 17 \cdot \int u^{4} d u

Apply the Power Rule

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

= - 17 \cdot \frac{u ^{5}}{5}

Substitute back

u = sin (t)

= - 17 \cdot \frac{sin ^{5} ( t )}{5}

Simplify

- 17 \cdot \frac{sin ^{5} ( t )}{5} : - \frac{17}{5} sin^{5} (t)

= - \frac{17}{5} sin^{5} (t)

Add a constant to the solution

= - \frac{17}{5} sin^{5} (t) + C

\frac{d}{d x} (\frac{x}{3 + x})

\frac{\partial}{\partial t} (cos (t))

x \to 0 lim (4 sin (x) ln (x))

(x^{2} + 1) y^{^{'}} = - 3 x (y - 1)

- \frac{1}{2 x ^{4}}

What is the integral of-17sin^4(t)cos(t) ?
The integral of-17sin^4(t)cos(t) is -17/5 sin^5(t)+C