What is the integral of sin^3(x/2)cos^7(x/2) ?

The integral of sin^3(x/2)cos^7(x/2) is 2(-(cos^8(x/2))/8+(cos^{10}(x/2))/(10))+C

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integral of sin^3(x/2)cos^7(x/2)

\int sin^{3} (\frac{x}{2}) cos^{7} (\frac{x}{2}) d x

2 (- \frac{cos ^{8} ( \frac{x}{2} )}{8} + \frac{cos ^{10} ( \frac{x}{2} )}{10}) + C

Solution steps

\int sin^{3} (\frac{x}{2}) cos^{7} (\frac{x}{2}) d x

Apply u-substitution

= \int sin^{3} (u) cos^{7} (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^{7} (u) d u

Rewrite using trig identities

= 2 \cdot \int (1 - cos^{2} (u)) sin (u) cos^{7} (u) d u

Apply u-substitution

= 2 \cdot \int - v^{7} (1 - v^{2}) d v

Expand

- v^{7} (1 - v^{2}) : - v^{7} + v^{9}

= 2 \cdot \int - v^{7} + v^{9} 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^{7} d v + \int v^{9} d v)

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

\int v^{9} d v = \frac{v ^{10}}{10}

= 2 (- \frac{v ^{8}}{8} + \frac{v ^{10}}{10})

Substitute back

= 2 (- \frac{cos ^{8} ( \frac{x}{2} )}{8} + \frac{cos ^{10} ( \frac{x}{2} )}{10})

Add a constant to the solution

= 2 (- \frac{cos ^{8} ( \frac{x}{2} )}{8} + \frac{cos ^{10} ( \frac{x}{2} )}{10}) + C

What is the integral of sin^3(x/2)cos^7(x/2) ?
The integral of sin^3(x/2)cos^7(x/2) is 2(-(cos^8(x/2))/8+(cos^{10}(x/2))/(10))+C