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

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

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

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

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

Solution steps

\int - 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

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

Apply u-substitution

: \int u^{2} d u

= - \int u^{2} d u

Apply the Power Rule

: \frac{u ^{3}}{3}

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

Substitute back

u = sin (t)

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{e ^{x}}{e ^{x} + 5})

\int 3 cos^{2} (x) sin (2 x) d x

\frac{d}{d x} (8 4 x - 1)

x \to 2 lim (x^{2} + a)

y^{^{'''}} - 3 y^{^{''}} + 3 y^{^{'}} - y = x

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