What is the integral of 5sin(5θ)sin(θ) ?

The integral of 5sin(5θ)sin(θ) is 5/2 (-1/6 sin(6θ)+1/4 sin(4θ))+C

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integral of 5sin(5θ)sin(θ)

\int 5 sin (5 θ) sin (θ) d θ

\frac{5}{2} (- \frac{1}{6} sin (6 θ) + \frac{1}{4} sin (4 θ)) + C

Solution steps

\int 5 sin (5 θ) sin (θ) d θ

Take the constant out:

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

= 5 \cdot \int sin (5 θ) sin (θ) d θ

Rewrite using trig identities

= 5 \cdot \int \frac{1}{2} (- cos (6 θ) + cos (4 θ)) d θ

Take the constant out:

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

= 5 \cdot \frac{1}{2} \cdot \int - cos (6 θ) + cos (4 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

= 5 \cdot \frac{1}{2} (- \int cos (6 θ) d θ + \int cos (4 θ) d θ)

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

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

= 5 \cdot \frac{1}{2} (- \frac{1}{6} sin (6 θ) + \frac{1}{4} sin (4 θ))

Simplify

5 \cdot \frac{1}{2} (- \frac{1}{6} sin (6 θ) + \frac{1}{4} sin (4 θ)) : \frac{5}{2} (- \frac{1}{6} sin (6 θ) + \frac{1}{4} sin (4 θ))

= \frac{5}{2} (- \frac{1}{6} sin (6 θ) + \frac{1}{4} sin (4 θ))

Add a constant to the solution

= \frac{5}{2} (- \frac{1}{6} sin (6 θ) + \frac{1}{4} sin (4 θ)) + C

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What is the integral of 5sin(5θ)sin(θ) ?
The integral of 5sin(5θ)sin(θ) is 5/2 (-1/6 sin(6θ)+1/4 sin(4θ))+C