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

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

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integral of sin(θ)cos(2θ)

\int sin (θ) cos (2 θ) d θ

\frac{1}{2} (- \frac{1}{3} cos (3 θ) + cos (θ)) + C

Solution steps

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

Simplify

sin (θ + 2 θ) + sin (θ - 2 θ) : sin (3 θ) - sin (θ)

= \frac{1}{2} \cdot \int sin (3 θ) - sin (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) d θ - \int sin (θ) d θ)

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

\int sin (θ) d θ = - cos (θ)

= \frac{1}{2} (- \frac{1}{3} cos (3 θ) - (- cos (θ)))

Simplify

= \frac{1}{2} (- \frac{1}{3} cos (3 θ) + cos (θ))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{3} cos (3 θ) + cos (θ)) + C

\int_{0}^{r} x arcsin (\frac{r - x}{r}) d x

\frac{d}{d x} (\frac{1}{sin ( 2 x )})

\frac{e ^{- s} - e ^{- 2 s}}{s ( s + 2 ) ( s + 1 )}

\int_{1}^{2} x 8^{x^{2}} d x

y^{^{'}} = \frac{y ( x - y )}{x ^{2}}

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