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

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

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

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

2 (\frac{sin ^{3} ( \frac{x}{2} )}{3} - \frac{sin ^{5} ( \frac{x}{2} )}{5}) + C

Solution steps

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

Apply u-substitution

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

= \int cos^{3} (u) sin^{2} (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 cos^{3} (u) sin^{2} (u) d u

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

v^{2} (1 - v^{2}) : v^{2} - v^{4}

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

\int v^{2} d v = \frac{v ^{3}}{3}

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

= 2 (\frac{v ^{3}}{3} - \frac{v ^{5}}{5})

Substitute back

= 2 (\frac{sin ^{3} ( \frac{x}{2} )}{3} - \frac{sin ^{5} ( \frac{x}{2} )}{5})

Add a constant to the solution

= 2 (\frac{sin ^{3} ( \frac{x}{2} )}{3} - \frac{sin ^{5} ( \frac{x}{2} )}{5}) + C

t^{2} + 1

y = sec (x) - 6 cos (x)

y = x^{8} + 7 x^{6} - 5 x + 4

\frac{d}{d x} (sin^{2} (x - 1))

\int (2 x^{2} - 1) (2 x^{3} - 3 x + 9)^{\frac{1}{3}} d x

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