What is the integral of (x(sin(x^2)))(cos^8(x^2)) ?

The integral of (x(sin(x^2)))(cos^8(x^2)) is -1/18 cos^9(x^2)+C

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integral of (x(sin(x^2)))(cos^8(x^2))

\int (x (sin (x^{2}))) (cos^{8} (x^{2})) d x

- \frac{1}{18} cos^{9} (x^{2}) + C

Solution steps

\int x sin (x^{2}) cos^{8} (x^{2}) d x

Apply u-substitution

: \int \frac{cos ^{8} ( u ) sin ( u )}{2} d u

= \int \frac{cos ^{8} ( u ) sin ( u )}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int cos^{8} (u) sin (u) d u

Apply u-substitution

: \int - v^{8} d v

= \frac{1}{2} \cdot \int - v^{8} d v

Take the constant out:

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

= \frac{1}{2} (- \int v^{8} d v)

Apply the Power Rule

: \frac{v ^{9}}{9}

= \frac{1}{2} (- \frac{v ^{9}}{9})

Substitute back

= \frac{1}{2} (- \frac{cos ^{9} ( x ^{2} )}{9})

Simplify

\frac{1}{2} (- \frac{cos ^{9} ( x ^{2} )}{9}) : - \frac{1}{18} cos^{9} (x^{2})

= - \frac{1}{18} cos^{9} (x^{2})

Add a constant to the solution

= - \frac{1}{18} cos^{9} (x^{2}) + C

\frac{d}{d x} (81 cos (3 x))

\frac{\partial}{\partial x} (x^{3} - 6 x^{2} - 34 x + 40)

f (x) = x (x - 3)

\frac{d}{d x} ((3 x - 1)^{4} (- 2 x + 9)^{5})

\frac{d}{d p} (\frac{400}{p ^{2}})

What is the integral of (x(sin(x^2)))(cos^8(x^2)) ?
The integral of (x(sin(x^2)))(cos^8(x^2)) is -1/18 cos^9(x^2)+C