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

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

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

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

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

Solution steps

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

Rewrite using trig identities

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

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

Add a constant to the solution

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

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x \to 0 lim ((1 - 3 x)^{\frac{2}{x}})

\int \frac{1}{( x ^{2} - 4 x + 5 ) ^{2}} d x

f (x) = 3 x^{2} + x - 2, at x = 1

\int \frac{x - 2}{x ^{2} - x + 1} d x

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