What is the integral of-2cos(t)sin^2(t) ?

The integral of-2cos(t)sin^2(t) is -2/3 sin^3(t)+C

integral of-2cos(t)sin^2(t)

\int - 2 cos (t) sin^{2} (t) d t

- \frac{2}{3} sin^{3} (t) + C

Solution steps

\int - 2 cos (t) sin^{2} (t) d t

Take the constant out:

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

= - 2 \cdot \int cos (t) sin^{2} (t) d t

Apply u-substitution

= - 2 \cdot \int u^{2} d u

Apply the Power Rule

= - 2 \cdot \frac{u ^{3}}{3}

Substitute back

u = sin (t)

= - 2 \cdot \frac{sin ^{3} ( t )}{3}

Simplify

- 2 \cdot \frac{sin ^{3} ( t )}{3} : - \frac{2}{3} sin^{3} (t)

= - \frac{2}{3} sin^{3} (t)

Add a constant to the solution

= - \frac{2}{3} sin^{3} (t) + C

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