What is the integral of sin(2x)sin(3x) ?

The integral of sin(2x)sin(3x) is 1/2 (sin(x)-1/5 sin(5x))+C

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integral of sin(2x)sin(3x)

\int sin (2 x) sin (3 x) d x

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

Solution steps

\int sin (2 x) sin (3 x) d x

Rewrite using trig identities

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

Simplify

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

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

\int cos (x) d x = sin (x)

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

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

Add a constant to the solution

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

\int \frac{7}{x + 1} d x

\int 2 (sec (x))^{2} d x

\frac{\partial ^{2}}{\partial y ^{2}} (y^{4} sin (6 x))

\int (2 x + 7)^{\frac{3}{2}} d x

f (x) = \frac{x}{5} + \frac{7}{x}

What is the integral of sin(2x)sin(3x) ?
The integral of sin(2x)sin(3x) is 1/2 (sin(x)-1/5 sin(5x))+C