What is the integral of cos(θ/9)cos(θ/5) ?

The integral of cos(θ/9)cos(θ/5) is 1/2 (45/4 sin((4θ)/(45))+45/14 sin((14θ)/(45)))+C

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integral of cos(θ/9)cos(θ/5)

\int cos (\frac{θ}{9}) cos (\frac{θ}{5}) d θ

\frac{1}{2} (\frac{45}{4} sin (\frac{4 θ}{45}) + \frac{45}{14} sin (\frac{14 θ}{45})) + C

Solution steps

\int cos (\frac{θ}{9}) cos (\frac{θ}{5}) d θ

Rewrite using trig identities

= \int \frac{1}{2} (cos (\frac{14 θ}{45}) + cos (- \frac{4 θ}{45})) 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 cos (\frac{14 θ}{45}) + cos (- \frac{4 θ}{45}) d θ

Simplify

cos (\frac{14 θ}{45}) + cos (- \frac{4 θ}{45}) : cos (\frac{4 θ}{45}) + cos (\frac{14 θ}{45})

= \frac{1}{2} \cdot \int cos (\frac{4 θ}{45}) + cos (\frac{14 θ}{45}) 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 cos (\frac{4 θ}{45}) d θ + \int cos (\frac{14 θ}{45}) d θ)

\int cos (\frac{4 θ}{45}) d θ = \frac{45}{4} sin (\frac{4 θ}{45})

\int cos (\frac{14 θ}{45}) d θ = \frac{45}{14} sin (\frac{14 θ}{45})

= \frac{1}{2} (\frac{45}{4} sin (\frac{4 θ}{45}) + \frac{45}{14} sin (\frac{14 θ}{45}))

Add a constant to the solution

= \frac{1}{2} (\frac{45}{4} sin (\frac{4 θ}{45}) + \frac{45}{14} sin (\frac{14 θ}{45})) + C

What is the integral of cos(θ/9)cos(θ/5) ?
The integral of cos(θ/9)cos(θ/5) is 1/2 (45/4 sin((4θ)/(45))+45/14 sin((14θ)/(45)))+C