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

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

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integral of sin(3θ)cos(2θ)

\int sin (3 θ) cos (2 θ) d θ

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

Solution steps

\int sin (3 θ) cos (2 θ) d θ

Use the following identity:

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

= \int \frac{sin ( 3 θ + 2 θ ) + sin ( 3 θ - 2 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 sin (3 θ + 2 θ) + sin (3 θ - 2 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 sin (3 θ + 2 θ) d θ + \int sin (3 θ - 2 θ) d θ)

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

\int sin (3 θ - 2 θ) d θ = - cos (θ)

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

Add a constant to the solution

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

\frac{d y}{d x} = \frac{3 sec ( y )}{( x + 5 ) ^{2}}

f (u) = \frac{u}{u + \frac{a}{u}}

y^{^{'}} = - x (y + 1), y (0) = 2

\frac{\partial}{\partial y} (2 x + y)

\int 2 x (4 x^{2} - 10)^{2} d x

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