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

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

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

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

\frac{2}{3} cos^{3} (x) + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int - u^{2} d u

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

Take the constant out:

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

= - 2 (- \int u^{2} d u)

Apply the Power Rule

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

= - 2 (- \frac{u ^{3}}{3})

Substitute back

u = cos (x)

= - 2 (- \frac{cos ^{3} ( x )}{3})

Simplify

- 2 (- \frac{cos ^{3} ( x )}{3}) : \frac{2}{3} cos^{3} (x)

= \frac{2}{3} cos^{3} (x)

Add a constant to the solution

= \frac{2}{3} cos^{3} (x) + C

\frac{d v}{d t} = 32 - \frac{4}{5} v

\frac{\partial}{\partial x} (x^{5} y^{5} + 4 x^{7} y)

\int \frac{1}{( x - 5 ) ^{\frac{3}{2}}} d x

\frac{d}{d x} (\frac{( 3 x - 2 ) ( 2 x - 3 )}{- 2 x ^{2} + 4 x})

x y^{^{'}} + (1 + 2 x) y = 10 e^{- 2 x}

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