What is the integral of-4sin^3(x)cos^5(x) ?

The integral of-4sin^3(x)cos^5(x) is -4(-(cos^6(x))/6+(cos^8(x))/8)+C

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integral of-4sin^3(x)cos^5(x)

\int - 4 sin^{3} (x) cos^{5} (x) d x

- 4 (- \frac{cos ^{6} ( x )}{6} + \frac{cos ^{8} ( x )}{8}) + C

Solution steps

\int - 4 sin^{3} (x) cos^{5} (x) d x

Take the constant out:

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

= - 4 \cdot \int sin^{3} (x) cos^{5} (x) d x

Rewrite using trig identities

= - 4 \cdot \int (1 - cos^{2} (x)) sin (x) cos^{5} (x) d x

Apply u-substitution

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

= - 4 \cdot \int - u^{5} (1 - u^{2}) d u

Expand

- u^{5} (1 - u^{2}) : - u^{5} + u^{7}

= - 4 \cdot \int - u^{5} + u^{7} 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

= - 4 (- \int u^{5} d u + \int u^{7} d u)

\int u^{5} d u = \frac{u ^{6}}{6}

\int u^{7} d u = \frac{u ^{8}}{8}

= - 4 (- \frac{u ^{6}}{6} + \frac{u ^{8}}{8})

Substitute back

u = cos (x)

= - 4 (- \frac{cos ^{6} ( x )}{6} + \frac{cos ^{8} ( x )}{8})

Add a constant to the solution

= - 4 (- \frac{cos ^{6} ( x )}{6} + \frac{cos ^{8} ( x )}{8}) + C

y = x^{2} - 4 x + 1, y = x + 1

x \to 0 lim (\frac{- x ^{- x} - 1}{x})

\frac{d}{d x} (\frac{e ^{- x}}{x - 5} + 1)

x \to 0 + lim ((1 + 2 x)^{\frac{4}{x}})

x^{3} \frac{d ^{3} y}{d x ^{3}} + 2 x^{2} \frac{d ^{2} y}{d x ^{2}} - x \frac{d y}{d x} + y = 12 x^{2}

What is the integral of-4sin^3(x)cos^5(x) ?
The integral of-4sin^3(x)cos^5(x) is -4(-(cos^6(x))/6+(cos^8(x))/8)+C