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

The integral of (cos^3(2x))/(sqrt(sin(2x))) is sqrt(sin(2x))-1/5 sin^{5/2}(2x)+C

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

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

sin (2 x) - \frac{1}{5} sin^{\frac{5}{2}} (2 x) + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Rewrite using trig identities

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

Apply u-substitution

: \int \frac{1 - v ^{2}}{v} d v

= \frac{1}{2} \cdot \int \frac{1 - v ^{2}}{v} d v

Expand

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

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

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 \frac{1}{v} d v - \int v^{\frac{3}{2}} d v)

\int \frac{1}{v} d v = 2 v

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

= \frac{1}{2} (2 v - \frac{2}{5} v^{\frac{5}{2}})

Substitute back

= \frac{1}{2} (2 sin (2 x) - \frac{2}{5} sin^{\frac{5}{2}} (2 x))

Simplify

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

= sin (2 x) - \frac{1}{5} sin^{\frac{5}{2}} (2 x)

Add a constant to the solution

= sin (2 x) - \frac{1}{5} sin^{\frac{5}{2}} (2 x) + C

\frac{π}{4}

\int sin^{3} (4 x) cos^{2} (4 x) d x

\frac{\partial}{\partial x} (4 - x^{2})

\frac{\partial}{\partial y} (x y + \frac{2}{x} + \frac{4}{y})

(- 34, 74), (- 33, 44)

What is the integral of (cos^3(2x))/(sqrt(sin(2x))) ?
The integral of (cos^3(2x))/(sqrt(sin(2x))) is sqrt(sin(2x))-1/5 sin^{5/2}(2x)+C