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

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

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integral of 9sin^{-3/2}(x)cos^3(x)

\int 9 sin^{- \frac{3}{2}} (x) cos^{3} (x) d x

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

Solution steps

\int 9 sin^{- \frac{3}{2}} (x) cos^{3} (x) d x

Take the constant out:

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

= 9 \cdot \int sin^{- \frac{3}{2}} (x) cos^{3} (x) d x

Rewrite using trig identities

= 9 \cdot \int (1 - sin^{2} (x)) cos (x) sin^{- \frac{3}{2}} (x) d x

Apply u-substitution

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

= 9 \cdot \int \frac{1 - u ^{2}}{u ^{\frac{3}{2}}} d u

Expand

\frac{1 - u ^{2}}{u ^{\frac{3}{2}}} : \frac{1}{u ^{\frac{3}{2}}} - u^{\frac{1}{2}}

= 9 \cdot \int \frac{1}{u ^{\frac{3}{2}}} - u^{\frac{1}{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

= 9 (\int \frac{1}{u ^{\frac{3}{2}}} d u - \int u^{\frac{1}{2}} d u)

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

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

= 9 (- \frac{2}{u} - \frac{2}{3} u^{\frac{3}{2}})

Substitute back

u = sin (x)

= 9 (- \frac{2}{sin ( x )} - \frac{2}{3} sin^{\frac{3}{2}} (x))

Add a constant to the solution

= 9 (- \frac{2}{sin ( x )} - \frac{2}{3} sin^{\frac{3}{2}} (x)) + C

x \to 1 lim (x - 8)

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

\frac{d y}{d x} = \frac{2 x}{y ( 1 + x ^{2} )}

f (x) = ln ((x - 1) (x + 1)^{2})

\int_{1}^{x^{3}} t^{2} d t

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