What is the integral of x^8cos(x^9) ?

The integral of x^8cos(x^9) is 1/9 sin(x^9)+C

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integral of x^8cos(x^9)

\int x^{8} cos (x^{9}) d x

\frac{1}{9} sin (x^{9}) + C

Solution steps

\int x^{8} cos (x^{9}) d x

Apply u-substitution

: \int \frac{cos ( u )}{9} d u

= \int \frac{cos ( u )}{9} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int cos (u) d u

Use the common integral:

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

= \frac{1}{9} sin (u)

Substitute back

u = x^{9}

= \frac{1}{9} sin (x^{9})

Add a constant to the solution

= \frac{1}{9} sin (x^{9}) + C

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What is the integral of x^8cos(x^9) ?
The integral of x^8cos(x^9) is 1/9 sin(x^9)+C