What is the integral of sin(4/5 x)sin(3/8 x) ?

The integral of sin(4/5 x)sin(3/8 x) is 20/799 (47sin((17x)/(40))-17sin((47x)/(40)))+C

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integral of sin(4/5 x)sin(3/8 x)

\int sin (\frac{4}{5} x) sin (\frac{3}{8} x) d x

\frac{20}{799} (47 sin (\frac{17 x}{40}) - 17 sin (\frac{47 x}{40})) + C

Solution steps

\int sin (\frac{4}{5} x) sin (\frac{3}{8} x) d x

Apply u-substitution

: \int \frac{5}{4} sin (\frac{15 u}{32}) sin (u) d u

= \int \frac{5}{4} sin (\frac{15 u}{32}) sin (u) d u

Take the constant out:

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

= \frac{5}{4} \cdot \int sin (\frac{15 u}{32}) sin (u) d u

Rewrite using trig identities

= \frac{5}{4} \cdot \int \frac{1}{2} (- cos (\frac{47 u}{32}) + cos (- \frac{17 u}{32})) d u

Take the constant out:

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

= \frac{5}{4} \cdot \frac{1}{2} \cdot \int - cos (\frac{47 u}{32}) + cos (- \frac{17 u}{32}) d u

Simplify

- cos (\frac{47 u}{32}) + cos (- \frac{17 u}{32}) : cos (\frac{17 u}{32}) - cos (\frac{47 u}{32})

= \frac{5}{4} \cdot \frac{1}{2} \cdot \int cos (\frac{17 u}{32}) - cos (\frac{47 u}{32}) d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{5}{4} \cdot \frac{1}{2} (\int cos (\frac{17 u}{32}) d u - \int cos (\frac{47 u}{32}) d u)

\int cos (\frac{17 u}{32}) d u = \frac{32}{17} sin (\frac{17 u}{32})

\int cos (\frac{47 u}{32}) d u = \frac{32}{47} sin (\frac{47 u}{32})

= \frac{5}{4} \cdot \frac{1}{2} (\frac{32}{17} sin (\frac{17 u}{32}) - \frac{32}{47} sin (\frac{47 u}{32}))

Substitute back

u = \frac{4}{5} x

= \frac{5}{4} \cdot \frac{1}{2} (\frac{32}{17} sin (\frac{17 \cdot \frac{4}{5} x}{32}) - \frac{32}{47} sin (\frac{47 \cdot \frac{4}{5} x}{32}))

Simplify

\frac{5}{4} \cdot \frac{1}{2} (\frac{32}{17} sin (\frac{17 \cdot \frac{4}{5} x}{32}) - \frac{32}{47} sin (\frac{47 \cdot \frac{4}{5} x}{32})) : \frac{20}{799} (47 sin (\frac{17 x}{40}) - 17 sin (\frac{47 x}{40}))

= \frac{20}{799} (47 sin (\frac{17 x}{40}) - 17 sin (\frac{47 x}{40}))

Add a constant to the solution

= \frac{20}{799} (47 sin (\frac{17 x}{40}) - 17 sin (\frac{47 x}{40})) + C

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What is the integral of sin(4/5 x)sin(3/8 x) ?
The integral of sin(4/5 x)sin(3/8 x) is 20/799 (47sin((17x)/(40))-17sin((47x)/(40)))+C