What is the integral of sin^7(x)cos^2(x) ?

The integral of sin^7(x)cos^2(x) is -(cos^3(x))/3+(3cos^5(x))/5-(3cos^7(x))/7+(cos^9(x))/9+C

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integral of sin^7(x)cos^2(x)

\int sin^{7} (x) cos^{2} (x) d x

- \frac{cos ^{3} ( x )}{3} + \frac{3 cos ^{5} ( x )}{5} - \frac{3 cos ^{7} ( x )}{7} + \frac{cos ^{9} ( x )}{9} + C

Solution steps

\int sin^{7} (x) cos^{2} (x) d x

Rewrite using trig identities

= \int (1 - cos^{2} (x))^{3} sin (x) cos^{2} (x) d x

Apply u-substitution

: \int - u^{2} (1 - u^{2})^{3} d u

= \int - u^{2} (1 - u^{2})^{3} d u

Expand

- u^{2} (1 - u^{2})^{3} : - u^{2} + 3 u^{4} - 3 u^{6} + u^{8}

= \int - u^{2} + 3 u^{4} - 3 u^{6} + u^{8} 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 3 u^{4} d u - \int 3 u^{6} d u + \int u^{8} d u

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

\int 3 u^{4} d u = \frac{3 u ^{5}}{5}

\int 3 u^{6} d u = \frac{3 u ^{7}}{7}

\int u^{8} d u = \frac{u ^{9}}{9}

= - \frac{u ^{3}}{3} + \frac{3 u ^{5}}{5} - \frac{3 u ^{7}}{7} + \frac{u ^{9}}{9}

Substitute back

u = cos (x)

= - \frac{cos ^{3} ( x )}{3} + \frac{3 cos ^{5} ( x )}{5} - \frac{3 cos ^{7} ( x )}{7} + \frac{cos ^{9} ( x )}{9}

Add a constant to the solution

= - \frac{cos ^{3} ( x )}{3} + \frac{3 cos ^{5} ( x )}{5} - \frac{3 cos ^{7} ( x )}{7} + \frac{cos ^{9} ( x )}{9} + C

y = lo g_{5} (x)

\frac{d}{d x} (lo g_{10} (2 x^{2} + 5 x + 2))

\frac{3 x - y}{x}

\frac{\partial}{\partial x} (x^{0.2} y^{0.4})

x \to 3 lim (\frac{3 - x}{x ^{2} - 9})

What is the integral of sin^7(x)cos^2(x) ?
The integral of sin^7(x)cos^2(x) is -(cos^3(x))/3+(3cos^5(x))/5-(3cos^7(x))/7+(cos^9(x))/9+C