What is the integral of xcos^3(x^2) ?

The integral of xcos^3(x^2) is 1/2 (sin(x^2)-(sin^3(x^2))/3)+C

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integral of xcos^3(x^2)

\int x cos^{3} (x^{2}) d x

\frac{1}{2} (sin (x^{2}) - \frac{sin ^{3} ( x ^{2} )}{3}) + C

Solution steps

\int x cos^{3} (x^{2}) d x

Apply u-substitution

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

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

Rewrite using trig identities

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

Apply u-substitution

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

= \frac{1}{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

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

\int 1 d v = v

\int v^{2} d v = \frac{v ^{3}}{3}

= \frac{1}{2} (v - \frac{v ^{3}}{3})

Substitute back

= \frac{1}{2} (sin (x^{2}) - \frac{sin ^{3} ( x ^{2} )}{3})

Add a constant to the solution

= \frac{1}{2} (sin (x^{2}) - \frac{sin ^{3} ( x ^{2} )}{3}) + C

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

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What is the integral of xcos^3(x^2) ?
The integral of xcos^3(x^2) is 1/2 (sin(x^2)-(sin^3(x^2))/3)+C