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

The integral of (sin(x))^5(cos(x))^4 is -(cos^5(x))/5+(2cos^7(x))/7-(cos^9(x))/9+C

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

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

- \frac{cos ^{5} ( x )}{5} + \frac{2 cos ^{7} ( x )}{7} - \frac{cos ^{9} ( x )}{9} + C

Solution steps

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

Simplify

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

Rewrite using trig identities

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

Apply u-substitution

: \int - u^{4} (1 - u^{2})^{2} d u

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

Expand

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

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

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

\int 2 u^{6} d u = \frac{2 u ^{7}}{7}

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

= - \frac{u ^{5}}{5} + \frac{2 u ^{7}}{7} - \frac{u ^{9}}{9}

Substitute back

u = cos (x)

= - \frac{cos ^{5} ( x )}{5} + \frac{2 cos ^{7} ( x )}{7} - \frac{cos ^{9} ( x )}{9}

Add a constant to the solution

= - \frac{cos ^{5} ( x )}{5} + \frac{2 cos ^{7} ( x )}{7} - \frac{cos ^{9} ( x )}{9} + C

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What is the integral of (sin(x))^5(cos(x))^4 ?
The integral of (sin(x))^5(cos(x))^4 is -(cos^5(x))/5+(2cos^7(x))/7-(cos^9(x))/9+C