What is the integral from 1 to 2 of sin(2θ)cos(3θ) ?

The integral from 1 to 2 of sin(2θ)cos(3θ) is 1/2 (1/5 (-cos(10)+cos(5))-cos(1)+cos(2))

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integral from 1 to 2 of sin(2θ)cos(3θ)

\int_{1}^{2} sin (2 θ) cos (3 θ) d θ

\frac{1}{2} (\frac{1}{5} (- cos (10) + cos (5)) - cos (1) + cos (2))

Solution steps

\int_{1}^{2} sin (2 θ) cos (3 θ) d θ

Use the following identity:

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

= \int_{1}^{2} \frac{sin ( 2 θ + 3 θ ) + sin ( 2 θ - 3 θ )}{2} d θ

Take the constant out:

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

= \frac{1}{2} \cdot \int_{1}^{2} sin (2 θ + 3 θ) + sin (2 θ - 3 θ) d θ

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}^{2} sin (2 θ + 3 θ) d θ + \int_{1}^{2} sin (2 θ - 3 θ) d θ)

\int_{1}^{2} sin (2 θ + 3 θ) d θ = \frac{1}{5} (- cos (10) + cos (5))

\int_{1}^{2} sin (2 θ - 3 θ) d θ = - cos (1) + cos (2)

= \frac{1}{2} (\frac{1}{5} (- cos (10) + cos (5)) - cos (1) + cos (2))

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What is the integral from 1 to 2 of sin(2θ)cos(3θ) ?
The integral from 1 to 2 of sin(2θ)cos(3θ) is 1/2 (1/5 (-cos(10)+cos(5))-cos(1)+cos(2))