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

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

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

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

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

Solution steps

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

Apply u-substitution

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

= \int \frac{3 cos ( 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 cos (u) d u

Use the common integral:

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

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

Substitute back

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

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

Add a constant to the solution

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

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