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

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

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integral of xsin(x^2)cos(3x^2)

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

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

Solution steps

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

Apply u-substitution

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

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

Use the following identity:

cos (t) sin (s) = \frac{sin ( s + t ) + sin ( s - t )}{2}

= \frac{1}{2} \cdot \int \frac{sin ( u + 3 u ) + sin ( u - 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 \frac{1}{2} \cdot \int sin (u + 3 u) + sin (u - 3 u) d u

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} \cdot \frac{1}{2} (\int sin (u + 3 u) d u + \int sin (u - 3 u) d u)

\int sin (u + 3 u) d u = - \frac{1}{4} cos (4 u)

\int sin (u - 3 u) d u = \frac{1}{2} cos (2 u)

= \frac{1}{2} \cdot \frac{1}{2} (- \frac{1}{4} cos (4 u) + \frac{1}{2} cos (2 u))

Substitute back

u = x^{2}

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

Simplify

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

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

Add a constant to the solution

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

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