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

The integral of sin^{3/2}(x)cos^3(x) is 2/5 sin^{5/2}(x)-2/9 sin^{9/2}(x)+C

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

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

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

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

: \int u^{\frac{3}{2}} (1 - u^{2}) d u

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

Expand

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

= \int u^{\frac{3}{2}} - u^{\frac{7}{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

= \int u^{\frac{3}{2}} d u - \int u^{\frac{7}{2}} d u

\int u^{\frac{3}{2}} d u = \frac{2}{5} u^{\frac{5}{2}}

\int u^{\frac{7}{2}} d u = \frac{2}{9} u^{\frac{9}{2}}

= \frac{2}{5} u^{\frac{5}{2}} - \frac{2}{9} u^{\frac{9}{2}}

Substitute back

u = sin (x)

= \frac{2}{5} sin^{\frac{5}{2}} (x) - \frac{2}{9} sin^{\frac{9}{2}} (x)

Add a constant to the solution

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

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What is the integral of sin^{3/2}(x)cos^3(x) ?
The integral of sin^{3/2}(x)cos^3(x) is 2/5 sin^{5/2}(x)-2/9 sin^{9/2}(x)+C