What is the integral from-2 to 4 of sin(8x)cos(15x) ?

The integral from-2 to 4 of sin(8x)cos(15x) is 1/2 (1/23 (-cos(92)+cos(46))-1/7 (-cos(28)+cos(14)))

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integral from-2 to 4 of sin(8x)cos(15x)

\int_{- 2}^{4} sin (8 x) cos (15 x) d x

\frac{1}{2} (\frac{1}{23} (- cos (92) + cos (46)) - \frac{1}{7} (- cos (28) + cos (14)))

Decimal Notation

- 0.07430 \dots

Solution steps

\int_{- 2}^{4} sin (8 x) cos (15 x) d x

Use the following identity:

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

= \int_{- 2}^{4} \frac{sin ( 8 x + 15 x ) + sin ( 8 x - 15 x )}{2} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int_{- 2}^{4} sin (8 x + 15 x) + sin (8 x - 15 x) d x

Simplify

sin (8 x + 15 x) + sin (8 x - 15 x) : sin (23 x) - sin (7 x)

= \frac{1}{2} \cdot \int_{- 2}^{4} sin (23 x) - sin (7 x) d x

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_{- 2}^{4} sin (23 x) d x - \int_{- 2}^{4} sin (7 x) d x)

\int_{- 2}^{4} sin (23 x) d x = \frac{1}{23} (- cos (92) + cos (46))

\int_{- 2}^{4} sin (7 x) d x = \frac{1}{7} (- cos (28) + cos (14))

= \frac{1}{2} (\frac{1}{23} (- cos (92) + cos (46)) - \frac{1}{7} (- cos (28) + cos (14)))

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What is the integral from-2 to 4 of sin(8x)cos(15x) ?
The integral from-2 to 4 of sin(8x)cos(15x) is 1/2 (1/23 (-cos(92)+cos(46))-1/7 (-cos(28)+cos(14)))