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

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

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

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

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

Solution steps

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

Rewrite using trig identities

= \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

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

Expand

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

= \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

= - \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}

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

Substitute back

u = cos (x)

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

Add a constant to the solution

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

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

\frac{\partial}{\partial x} (2 tan (x y))

y^{^{'}} + a y = 0

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

n = 1 \sum \infty \frac{( 1 )}{n !}

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