What is the integral of (7sin^3(sqrt(x)))/(sqrt(x)) ?

The integral of (7sin^3(sqrt(x)))/(sqrt(x)) is 14(-cos(sqrt(x))+1/3 cos^3(sqrt(x)))+C

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integral of (7sin^3(sqrt(x)))/(sqrt(x))

\int \frac{7 sin ^{3} ( x )}{x} d x

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

Solution steps

\int \frac{7 sin ^{3} ( x )}{x} d x

Take the constant out:

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

= 7 \cdot \int \frac{sin ^{3} ( x )}{x} d x

Apply u-substitution

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

= 7 \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

= 7 \cdot 2 \cdot \int sin^{3} (u) d u

Rewrite using trig identities

= 7 \cdot 2 \cdot \int (1 - cos^{2} (u)) sin (u) d u

Apply u-substitution

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

= 7 \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

= 7 \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}

= 7 \cdot 2 (- v + \frac{v ^{3}}{3})

Substitute back

= 7 \cdot 2 (- cos (x) + \frac{cos ^{3} ( x )}{3})

Simplify

7 \cdot 2 (- cos (x) + \frac{cos ^{3} ( x )}{3}) : 14 (- cos (x) + \frac{1}{3} cos^{3} (x))

= 14 (- cos (x) + \frac{1}{3} cos^{3} (x))

Add a constant to the solution

= 14 (- cos (x) + \frac{1}{3} cos^{3} (x)) + C

X^{2}

\int (\frac{1}{x ^{2}} + \frac{1}{x x} + 5) d x

x \to 5 - lim (\frac{( x + 1 )}{x - 5})

\int (x^{2} + 1) e^{- x} d x

\frac{d}{d x} (- sin (x) - 2 cos (x))

What is the integral of (7sin^3(sqrt(x)))/(sqrt(x)) ?
The integral of (7sin^3(sqrt(x)))/(sqrt(x)) is 14(-cos(sqrt(x))+1/3 cos^3(sqrt(x)))+C