What is the integral of sin^5(x/2)cos^4(x/2) ?

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

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integral of sin^5(x/2)cos^4(x/2)

\int sin^{5} (\frac{x}{2}) cos^{4} (\frac{x}{2}) d x

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

Solution steps

\int sin^{5} (\frac{x}{2}) cos^{4} (\frac{x}{2}) d x

Apply u-substitution

: \int sin^{5} (u) cos^{4} (u) \cdot 2 d u

= \int sin^{5} (u) cos^{4} (u) \cdot 2 d u

Take the constant out:

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

= 2 \cdot \int sin^{5} (u) cos^{4} (u) d u

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

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

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

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 v^{4} d v + \int 2 v^{6} d v - \int v^{8} d v)

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

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

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

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

Substitute back

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

Add a constant to the solution

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

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