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

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

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

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

- cos (x) + \frac{1}{3} cos^{3} (x) + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

Rewrite using trig identities

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

Apply u-substitution

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

\int 1 d v = v

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

= \frac{1}{2} \cdot 2 (- v + \frac{v ^{3}}{3})

Substitute back

= \frac{1}{2} \cdot 2 (- cos (x) + \frac{cos ^{3} ( x )}{3})

Simplify

\frac{1}{2} \cdot 2 (- cos (x) + \frac{cos ^{3} ( x )}{3}) : - cos (x) + \frac{1}{3} cos^{3} (x)

= - cos (x) + \frac{1}{3} cos^{3} (x)

Add a constant to the solution

= - cos (x) + \frac{1}{3} cos^{3} (x) + C

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