What is the integral of sin^3(x/2)*cos^5(x/2) ?

The integral of sin^3(x/2)*cos^5(x/2) is 2(-(cos^6(x/2))/6+(cos^8(x/2))/8)+C

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integral of sin^3(x/2)*cos^5(x/2)

\int sin^{3} (\frac{x}{2}) \cdot cos^{5} (\frac{x}{2}) d x

2 (- \frac{cos ^{6} ( \frac{x}{2} )}{6} + \frac{cos ^{8} ( \frac{x}{2} )}{8}) + C

Solution steps

\int sin^{3} (\frac{x}{2}) cos^{5} (\frac{x}{2}) d x

Apply u-substitution

: \int sin^{3} (u) cos^{5} (u) \cdot 2 d u

= \int sin^{3} (u) cos^{5} (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^{3} (u) cos^{5} (u) d u

Rewrite using trig identities

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

Apply u-substitution

: \int - v^{5} (1 - v^{2}) d v

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

Expand

- v^{5} (1 - v^{2}) : - v^{5} + v^{7}

= 2 \cdot \int - v^{5} + v^{7} 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^{5} d v + \int v^{7} d v)

\int v^{5} d v = \frac{v ^{6}}{6}

\int v^{7} d v = \frac{v ^{8}}{8}

= 2 (- \frac{v ^{6}}{6} + \frac{v ^{8}}{8})

Substitute back

= 2 (- \frac{cos ^{6} ( \frac{x}{2} )}{6} + \frac{cos ^{8} ( \frac{x}{2} )}{8})

Add a constant to the solution

= 2 (- \frac{cos ^{6} ( \frac{x}{2} )}{6} + \frac{cos ^{8} ( \frac{x}{2} )}{8}) + C

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What is the integral of sin^3(x/2)*cos^5(x/2) ?
The integral of sin^3(x/2)*cos^5(x/2) is 2(-(cos^6(x/2))/6+(cos^8(x/2))/8)+C