What is the integral of-5sin^3(x)cos^2(x) ?

The integral of-5sin^3(x)cos^2(x) is 5/3 cos^3(x)-cos^5(x)+C

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integral of-5sin^3(x)cos^2(x)

\int - 5 sin^{3} (x) cos^{2} (x) d x

\frac{5}{3} cos^{3} (x) - cos^{5} (x) + C

Solution steps

\int - 5 sin^{3} (x) cos^{2} (x) d x

Take the constant out:

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

= - 5 \cdot \int sin^{3} (x) cos^{2} (x) d x

Rewrite using trig identities

= - 5 \cdot \int (1 - cos^{2} (x)) sin (x) cos^{2} (x) d x

Apply u-substitution

: \int - u^{2} (1 - u^{2}) d u

= - 5 \cdot \int - u^{2} (1 - u^{2}) d u

Expand

- u^{2} (1 - u^{2}) : - u^{2} + u^{4}

= - 5 \cdot \int - u^{2} + u^{4} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - 5 (- \int u^{2} d u + \int u^{4} d u)

\int u^{2} d u = \frac{u ^{3}}{3}

\int u^{4} d u = \frac{u ^{5}}{5}

= - 5 (- \frac{u ^{3}}{3} + \frac{u ^{5}}{5})

Substitute back

u = cos (x)

= - 5 (- \frac{cos ^{3} ( x )}{3} + \frac{cos ^{5} ( x )}{5})

Simplify

- 5 (- \frac{cos ^{3} ( x )}{3} + \frac{cos ^{5} ( x )}{5}) : \frac{5}{3} cos^{3} (x) - cos^{5} (x)

= \frac{5}{3} cos^{3} (x) - cos^{5} (x)

Add a constant to the solution

= \frac{5}{3} cos^{3} (x) - cos^{5} (x) + C

\frac{d y}{d t} = 1 0^{- 21} \cdot y^{3}

f (x) = - 3 x^{2} + 8 x + 1, at x = - 2

3^{- x}

x \to 6 lim (- x^{2} + 4 x - 3)

\int csc (x) sec^{2} (x) d x

What is the integral of-5sin^3(x)cos^2(x) ?
The integral of-5sin^3(x)cos^2(x) is 5/3 cos^3(x)-cos^5(x)+C