What is the integral of x^{1/3}sin(x^{4/3}-8) ?

The integral of x^{1/3}sin(x^{4/3}-8) is -3/4 cos(x^{4/3}-8)+C

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integral of x^{1/3}sin(x^{4/3}-8)

\int x^{\frac{1}{3}} sin (x^{\frac{4}{3}} - 8) d x

- \frac{3}{4} cos (x^{\frac{4}{3}} - 8) + C

Solution steps

\int x^{\frac{1}{3}} sin (x^{\frac{4}{3}} - 8) d x

Apply u-substitution

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

Take the constant out:

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

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

Use the common integral:

\int sin (u) d u = - cos (u)

= \frac{3}{4} (- cos (u))

Substitute back

u = x^{\frac{4}{3}} - 8

= \frac{3}{4} (- cos (x^{\frac{4}{3}} - 8))

Simplify

= - \frac{3}{4} cos (x^{\frac{4}{3}} - 8)

Add a constant to the solution

= - \frac{3}{4} cos (x^{\frac{4}{3}} - 8) + C

What is the integral of x^{1/3}sin(x^{4/3}-8) ?
The integral of x^{1/3}sin(x^{4/3}-8) is -3/4 cos(x^{4/3}-8)+C