What is the integral of sin(2x)sin(8x) ?

The integral of sin(2x)sin(8x) is 1/2 (1/6 sin(6x)-1/10 sin(10x))+C

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integral of sin(2x)sin(8x)

\int sin (2 x) sin (8 x) d x

\frac{1}{2} (\frac{1}{6} sin (6 x) - \frac{1}{10} sin (10 x)) + C

Solution steps

\int sin (2 x) sin (8 x) d x

Rewrite using trig identities

= \int \frac{1}{2} (- cos (10 x) + cos (- 6 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 (10 x) + cos (- 6 x) d x

Simplify

- cos (10 x) + cos (- 6 x) : cos (6 x) - cos (10 x)

= \frac{1}{2} \cdot \int cos (6 x) - cos (10 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 (6 x) d x - \int cos (10 x) d x)

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

\int cos (10 x) d x = \frac{1}{10} sin (10 x)

= \frac{1}{2} (\frac{1}{6} sin (6 x) - \frac{1}{10} sin (10 x))

Add a constant to the solution

= \frac{1}{2} (\frac{1}{6} sin (6 x) - \frac{1}{10} sin (10 x)) + C

y^{^{''}} + 5 y = 0, y^{^{'}} (1) = 3

x \to 0 + lim (tan (x))

\int \frac{1}{cos ( 2 y ) - cos ^{2} ( y )} d y

f (x) = (x - 2) (x^{2} + 4), (1, - 5)

sin (x), 0, π

What is the integral of sin(2x)sin(8x) ?
The integral of sin(2x)sin(8x) is 1/2 (1/6 sin(6x)-1/10 sin(10x))+C