What is the integral of cos(θ/9)cos(θ/4) ?

The integral of cos(θ/9)cos(θ/4) is 1/2 (36/5 sin((5θ)/(36))+36/13 sin((13θ)/(36)))+C

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integral of cos(θ/9)cos(θ/4)

\int cos (\frac{θ}{9}) cos (\frac{θ}{4}) d θ

\frac{1}{2} (\frac{36}{5} sin (\frac{5 θ}{36}) + \frac{36}{13} sin (\frac{13 θ}{36})) + C

Solution steps

\int cos (\frac{θ}{9}) cos (\frac{θ}{4}) d θ

Rewrite using trig identities

= \int \frac{1}{2} (cos (\frac{13 θ}{36}) + cos (- \frac{5 θ}{36})) 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{13 θ}{36}) + cos (- \frac{5 θ}{36}) d θ

Simplify

cos (\frac{13 θ}{36}) + cos (- \frac{5 θ}{36}) : cos (\frac{5 θ}{36}) + cos (\frac{13 θ}{36})

= \frac{1}{2} \cdot \int cos (\frac{5 θ}{36}) + cos (\frac{13 θ}{36}) 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{5 θ}{36}) d θ + \int cos (\frac{13 θ}{36}) d θ)

\int cos (\frac{5 θ}{36}) d θ = \frac{36}{5} sin (\frac{5 θ}{36})

\int cos (\frac{13 θ}{36}) d θ = \frac{36}{13} sin (\frac{13 θ}{36})

= \frac{1}{2} (\frac{36}{5} sin (\frac{5 θ}{36}) + \frac{36}{13} sin (\frac{13 θ}{36}))

Add a constant to the solution

= \frac{1}{2} (\frac{36}{5} sin (\frac{5 θ}{36}) + \frac{36}{13} sin (\frac{13 θ}{36})) + C

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What is the integral of cos(θ/9)cos(θ/4) ?
The integral of cos(θ/9)cos(θ/4) is 1/2 (36/5 sin((5θ)/(36))+36/13 sin((13θ)/(36)))+C