What is the integral of sin(x)sin(6x) ?

The integral of sin(x)sin(6x) is 1/2 (1/5 sin(5x)-1/7 sin(7x))+C

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integral of sin(x)sin(6x)

\int sin (x) sin (6 x) d x

\frac{1}{2} (\frac{1}{5} sin (5 x) - \frac{1}{7} sin (7 x)) + C

Solution steps

\int sin (x) sin (6 x) d x

Rewrite using trig identities

= \int \frac{1}{2} (- cos (7 x) + cos (- 5 x)) d x

Take the constant out:

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

= \frac{1}{2} \cdot \int - cos (7 x) + cos (- 5 x) d x

Simplify

- cos (7 x) + cos (- 5 x) : cos (5 x) - cos (7 x)

= \frac{1}{2} \cdot \int cos (5 x) - cos (7 x) d x

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 (5 x) d x - \int cos (7 x) d x)

\int cos (5 x) d x = \frac{1}{5} sin (5 x)

\int cos (7 x) d x = \frac{1}{7} sin (7 x)

= \frac{1}{2} (\frac{1}{5} sin (5 x) - \frac{1}{7} sin (7 x))

Add a constant to the solution

= \frac{1}{2} (\frac{1}{5} sin (5 x) - \frac{1}{7} sin (7 x)) + C

\frac{1}{3 x ^{\frac{2}{3}}}

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

\int_{- 32}^{8} \frac{1}{∣ 2 x ∣} d x

63 cos (7 x)

\frac{1}{4} x^{2} - 1, y = 8, y = 2

What is the integral of sin(x)sin(6x) ?
The integral of sin(x)sin(6x) is 1/2 (1/5 sin(5x)-1/7 sin(7x))+C