What is the integral of (cos^3(4x))/(sqrt(sin(4x))) ?

The integral of (cos^3(4x))/(sqrt(sin(4x))) is 1/4 (2sqrt(sin(4x))-2/5 sin^{5/2}(4x))+C

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integral of (cos^3(4x))/(sqrt(sin(4x)))

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

\frac{1}{4} (2 sin (4 x) - \frac{2}{5} sin^{\frac{5}{2}} (4 x)) + C

Solution steps

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

Apply u-substitution

= \int \frac{cos ^{3} ( u )}{4 sin ( u )} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{cos ^{3} ( u )}{sin ( u )} d u

Rewrite using trig identities

= \frac{1}{4} \cdot \int \frac{( 1 - sin ^{2} ( u ) ) cos ( u )}{sin ( u )} d u

Apply u-substitution

= \frac{1}{4} \cdot \int \frac{1 - v ^{2}}{v} d v

Expand

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

= \frac{1}{4} \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}{4} (\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}{4} (2 v - \frac{2}{5} v^{\frac{5}{2}})

Substitute back

= \frac{1}{4} (2 sin (4 x) - \frac{2}{5} sin^{\frac{5}{2}} (4 x))

Add a constant to the solution

= \frac{1}{4} (2 sin (4 x) - \frac{2}{5} sin^{\frac{5}{2}} (4 x)) + C

3 x^{2} y^{2} d x + (2 y^{3} + 4 y^{3}) d y = 0

\frac{\partial}{\partial x} (x^{7} y^{5} - x^{6} y^{4})

in v erse l a pl a ce \frac{3}{5 s - 2}

\int \frac{2 x + 1}{x ^{3} - x ^{2}} d x

\frac{\partial}{\partial x} (- x y z^{2} + 6)

What is the integral of (cos^3(4x))/(sqrt(sin(4x))) ?
The integral of (cos^3(4x))/(sqrt(sin(4x))) is 1/4 (2sqrt(sin(4x))-2/5 sin^{5/2}(4x))+C