What is the integral of 4xcos^3(x^2)sin(x^2) ?

The integral of 4xcos^3(x^2)sin(x^2) is -1/2 cos^4(x^2)+C

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integral of 4xcos^3(x^2)sin(x^2)

\int 4 x cos^{3} (x^{2}) sin (x^{2}) d x

- \frac{1}{2} cos^{4} (x^{2}) + C

Solution steps

\int 4 x cos^{3} (x^{2}) sin (x^{2}) d x

Take the constant out:

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

= 4 \cdot \int x cos^{3} (x^{2}) sin (x^{2}) d x

Apply u-substitution

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

= 4 \cdot \int \frac{cos ^{3} ( u ) sin ( u )}{2} d u

Take the constant out:

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

= 4 \cdot \frac{1}{2} \cdot \int cos^{3} (u) sin (u) d u

Apply u-substitution

: \int - v^{3} d v

= 4 \cdot \frac{1}{2} \cdot \int - v^{3} d v

Take the constant out:

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

= 4 \cdot \frac{1}{2} (- \int v^{3} d v)

Apply the Power Rule

: \frac{v ^{4}}{4}

= 4 \cdot \frac{1}{2} (- \frac{v ^{4}}{4})

Substitute back

= 4 \cdot \frac{1}{2} (- \frac{cos ^{4} ( x ^{2} )}{4})

Simplify

4 \cdot \frac{1}{2} (- \frac{cos ^{4} ( x ^{2} )}{4}) : - \frac{1}{2} cos^{4} (x^{2})

= - \frac{1}{2} cos^{4} (x^{2})

Add a constant to the solution

= - \frac{1}{2} cos^{4} (x^{2}) + C

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What is the integral of 4xcos^3(x^2)sin(x^2) ?
The integral of 4xcos^3(x^2)sin(x^2) is -1/2 cos^4(x^2)+C