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

The integral of 5sin^5(x)cos^4(x) is -cos^5(x)+10/7 cos^7(x)-5/9 cos^9(x)+C

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

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

- cos^{5} (x) + \frac{10}{7} cos^{7} (x) - \frac{5}{9} cos^{9} (x) + C

Solution steps

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

Take the constant out:

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

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

Rewrite using trig identities

= 5 \cdot \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

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

Expand

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

= 5 \cdot \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

= 5 (- \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}

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

Substitute back

u = cos (x)

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

Simplify

5 (- \frac{cos ^{5} ( x )}{5} + \frac{2 cos ^{7} ( x )}{7} - \frac{cos ^{9} ( x )}{9}) : - cos^{5} (x) + \frac{10}{7} cos^{7} (x) - \frac{5}{9} cos^{9} (x)

= - cos^{5} (x) + \frac{10}{7} cos^{7} (x) - \frac{5}{9} cos^{9} (x)

Add a constant to the solution

= - cos^{5} (x) + \frac{10}{7} cos^{7} (x) - \frac{5}{9} cos^{9} (x) + C

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What is the integral of 5sin^5(x)cos^4(x) ?
The integral of 5sin^5(x)cos^4(x) is -cos^5(x)+10/7 cos^7(x)-5/9 cos^9(x)+C