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

The integral of 2sin^5(x)cos^{-3/2}(x) is 2(-2/5 cos^{5/2}(x)+4/9 cos^{9/2}(x)-2/13 cos^{13/2}(x))+C

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integral of 2sin^5(x)cos^{-3/2}(x)

\int 2 sin^{5} (x) cos^{- \frac{3}{2}} (x) d x

2 (- \frac{2}{5} cos^{\frac{5}{2}} (x) + \frac{4}{9} cos^{\frac{9}{2}} (x) - \frac{2}{13} cos^{\frac{13}{2}} (x)) + C

Solution steps

\int 2 sin^{5} (x) cos^{- \frac{3}{2}} (x) d x

Take the constant out:

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

= 2 \cdot \int sin^{5} (x) cos^{- \frac{3}{2}} (x) d x

- \frac{3}{2} = 3 \cdot \frac{1}{2}

= 2 \cdot \int sin^{5} (x) cos^{3 \cdot \frac{1}{2}} (x) d x

Multiply

3 \cdot \frac{1}{2} : \frac{3}{2}

= 2 \cdot \int sin^{5} (x) cos^{\frac{3}{2}} (x) d x

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

- u^{\frac{3}{2}} (1 - u^{2})^{2} : - u^{\frac{3}{2}} + 2 u^{\frac{7}{2}} - u^{\frac{11}{2}}

= 2 \cdot \int - u^{\frac{3}{2}} + 2 u^{\frac{7}{2}} - u^{\frac{11}{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

= 2 (- \int u^{\frac{3}{2}} d u + \int 2 u^{\frac{7}{2}} d u - \int u^{\frac{11}{2}} d u)

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

\int 2 u^{\frac{7}{2}} d u = \frac{4}{9} u^{\frac{9}{2}}

\int u^{\frac{11}{2}} d u = \frac{2}{13} u^{\frac{13}{2}}

= 2 (- \frac{2}{5} u^{\frac{5}{2}} + \frac{4}{9} u^{\frac{9}{2}} - \frac{2}{13} u^{\frac{13}{2}})

Substitute back

u = cos (x)

= 2 (- \frac{2}{5} cos^{\frac{5}{2}} (x) + \frac{4}{9} cos^{\frac{9}{2}} (x) - \frac{2}{13} cos^{\frac{13}{2}} (x))

Add a constant to the solution

= 2 (- \frac{2}{5} cos^{\frac{5}{2}} (x) + \frac{4}{9} cos^{\frac{9}{2}} (x) - \frac{2}{13} cos^{\frac{13}{2}} (x)) + C

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What is the integral of 2sin^5(x)cos^{-3/2}(x) ?
The integral of 2sin^5(x)cos^{-3/2}(x) is 2(-2/5 cos^{5/2}(x)+4/9 cos^{9/2}(x)-2/13 cos^{13/2}(x))+C