What is the integral of xsin(7x^2)cos(8x^2) ?

The integral of xsin(7x^2)cos(8x^2) is 1/4 (-1/15 cos(15x^2)+cos(x^2))+C

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integral of xsin(7x^2)cos(8x^2)

\int x sin (7 x^{2}) cos (8 x^{2}) d x

\frac{1}{4} (- \frac{1}{15} cos (15 x^{2}) + cos (x^{2})) + C

Solution steps

\int x sin (7 x^{2}) cos (8 x^{2}) d x

Apply u-substitution

: \int \frac{sin ( 7 u ) cos ( 8 u )}{2} d u

= \int \frac{sin ( 7 u ) cos ( 8 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 (7 u) cos (8 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 ( 7 u + 8 u ) + sin ( 7 u - 8 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 (7 u + 8 u) + sin (7 u - 8 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 (7 u + 8 u) d u + \int sin (7 u - 8 u) d u)

\int sin (7 u + 8 u) d u = - \frac{1}{15} cos (15 u)

\int sin (7 u - 8 u) d u = cos (u)

= \frac{1}{2} \cdot \frac{1}{2} (- \frac{1}{15} cos (15 u) + cos (u))

Substitute back

u = x^{2}

= \frac{1}{2} \cdot \frac{1}{2} (- \frac{1}{15} cos (15 x^{2}) + cos (x^{2}))

Simplify

\frac{1}{2} \cdot \frac{1}{2} (- \frac{1}{15} cos (15 x^{2}) + cos (x^{2})) : \frac{1}{4} (- \frac{1}{15} cos (15 x^{2}) + cos (x^{2}))

= \frac{1}{4} (- \frac{1}{15} cos (15 x^{2}) + cos (x^{2}))

Add a constant to the solution

= \frac{1}{4} (- \frac{1}{15} cos (15 x^{2}) + cos (x^{2})) + C

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What is the integral of xsin(7x^2)cos(8x^2) ?
The integral of xsin(7x^2)cos(8x^2) is 1/4 (-1/15 cos(15x^2)+cos(x^2))+C