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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Rewrite using trig identities

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

Apply u-substitution

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

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

Add a constant to the solution

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

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