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

The integral of-2sin^5(x)cos^2(x) is -2(-(cos^3(x))/3+(2cos^5(x))/5-(cos^7(x))/7)+C

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

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

- 2 (- \frac{cos ^{3} ( x )}{3} + \frac{2 cos ^{5} ( x )}{5} - \frac{cos ^{7} ( x )}{7}) + C

Solution steps

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

Take the constant out:

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

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

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

- u^{2} (1 - u^{2})^{2} : - u^{2} + 2 u^{4} - u^{6}

= - 2 \cdot \int - u^{2} + 2 u^{4} - u^{6} 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

= - 2 (- \int u^{2} d u + \int 2 u^{4} d u - \int u^{6} d u)

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

\int 2 u^{4} d u = \frac{2 u ^{5}}{5}

\int u^{6} d u = \frac{u ^{7}}{7}

= - 2 (- \frac{u ^{3}}{3} + \frac{2 u ^{5}}{5} - \frac{u ^{7}}{7})

Substitute back

u = cos (x)

= - 2 (- \frac{cos ^{3} ( x )}{3} + \frac{2 cos ^{5} ( x )}{5} - \frac{cos ^{7} ( x )}{7})

Add a constant to the solution

= - 2 (- \frac{cos ^{3} ( x )}{3} + \frac{2 cos ^{5} ( x )}{5} - \frac{cos ^{7} ( x )}{7}) + C

\frac{d}{d x} (1.795 x^{2})

t^{3} \frac{d x}{d t} + 3 t^{2} x = t

y = x^{2} e^{- x}, (1, \frac{1}{e})

\frac{d}{d x} (3^{x})

\int \frac{1}{x ^{2} - 289} d x

What is the integral of-2sin^5(x)cos^2(x) ?
The integral of-2sin^5(x)cos^2(x) is -2(-(cos^3(x))/3+(2cos^5(x))/5-(cos^7(x))/7)+C