What is the integral of sin(6θ)cos(2θ) ?

The integral of sin(6θ)cos(2θ) is 1/2 (-1/8 cos(8θ)-1/4 cos(4θ))+C

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integral of sin(6θ)cos(2θ)

\int sin (6 θ) cos (2 θ) d θ

\frac{1}{2} (- \frac{1}{8} cos (8 θ) - \frac{1}{4} cos (4 θ)) + C

Solution steps

\int sin (6 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 ( 6 θ + 2 θ ) + sin ( 6 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 (6 θ + 2 θ) + sin (6 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 (6 θ + 2 θ) d θ + \int sin (6 θ - 2 θ) d θ)

\int sin (6 θ + 2 θ) d θ = - \frac{1}{8} cos (8 θ)

\int sin (6 θ - 2 θ) d θ = - \frac{1}{4} cos (4 θ)

= \frac{1}{2} (- \frac{1}{8} cos (8 θ) - \frac{1}{4} cos (4 θ))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{8} cos (8 θ) - \frac{1}{4} cos (4 θ)) + C

f (x) = (\frac{2 x}{x ^{2} + 1}), at x = 1

y^{^{''}} = - 3 y

\frac{\partial}{\partial x} (sec (x y) + tan (x y))

x \to 4 + lim (\frac{2}{x - 4})

\int 3 x^{- 5} d x

What is the integral of sin(6θ)cos(2θ) ?
The integral of sin(6θ)cos(2θ) is 1/2 (-1/8 cos(8θ)-1/4 cos(4θ))+C