What is the integral of 1/2 cos^3(x)sqrt(sin(x)) ?

The integral of 1/2 cos^3(x)sqrt(sin(x)) is 1/3 sin^{3/2}(x)-1/7 sin^{7/2}(x)+C

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integral of 1/2 cos^3(x)sqrt(sin(x))

\int \frac{1}{2} cos^{3} (x) sin (x) d x

\frac{1}{3} sin^{\frac{3}{2}} (x) - \frac{1}{7} sin^{\frac{7}{2}} (x) + C

Solution steps

\int \frac{1}{2} cos^{3} (x) sin (x) d x

Take the constant out:

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

= \frac{1}{2} \cdot \int cos^{3} (x) sin (x) d x

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

u (1 - u^{2}) : u - u^{\frac{5}{2}}

= \frac{1}{2} \cdot \int u - u^{\frac{5}{2}} 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

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

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

\int u^{\frac{5}{2}} d u = \frac{2}{7} u^{\frac{7}{2}}

= \frac{1}{2} (\frac{2}{3} u^{\frac{3}{2}} - \frac{2}{7} u^{\frac{7}{2}})

Substitute back

u = sin (x)

= \frac{1}{2} (\frac{2}{3} sin^{\frac{3}{2}} (x) - \frac{2}{7} sin^{\frac{7}{2}} (x))

Simplify

\frac{1}{2} (\frac{2}{3} sin^{\frac{3}{2}} (x) - \frac{2}{7} sin^{\frac{7}{2}} (x)) : \frac{1}{3} sin^{\frac{3}{2}} (x) - \frac{1}{7} sin^{\frac{7}{2}} (x)

= \frac{1}{3} sin^{\frac{3}{2}} (x) - \frac{1}{7} sin^{\frac{7}{2}} (x)

Add a constant to the solution

= \frac{1}{3} sin^{\frac{3}{2}} (x) - \frac{1}{7} sin^{\frac{7}{2}} (x) + C

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What is the integral of 1/2 cos^3(x)sqrt(sin(x)) ?
The integral of 1/2 cos^3(x)sqrt(sin(x)) is 1/3 sin^{3/2}(x)-1/7 sin^{7/2}(x)+C