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

The integral of sin(x)sin(3x) is 1/2 (1/2 sin(2x)-1/4 sin(4x))+C

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integral of sin(x)sin(3x)

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

\frac{1}{2} (\frac{1}{2} sin (2 x) - \frac{1}{4} sin (4 x)) + C

Solution steps

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

Rewrite using trig identities

= \int \frac{1}{2} (- cos (4 x) + cos (- 2 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 (4 x) + cos (- 2 x) d x

Simplify

- cos (4 x) + cos (- 2 x) : cos (2 x) - cos (4 x)

= \frac{1}{2} \cdot \int cos (2 x) - cos (4 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 (2 x) d x - \int cos (4 x) d x)

\int cos (2 x) d x = \frac{1}{2} sin (2 x)

\int cos (4 x) d x = \frac{1}{4} sin (4 x)

= \frac{1}{2} (\frac{1}{2} sin (2 x) - \frac{1}{4} sin (4 x))

Add a constant to the solution

= \frac{1}{2} (\frac{1}{2} sin (2 x) - \frac{1}{4} sin (4 x)) + C

\frac{d y}{d x}, y = x^{3 x}

f (x) = t

\frac{d}{d x} (\frac{1}{x ^{2}} + \frac{1}{x ^{6}})

\frac{d}{d x} (\frac{3 x ^{4}}{5})

\frac{s}{s ^{2} + 6 s + 10}

What is the integral of sin(x)sin(3x) ?
The integral of sin(x)sin(3x) is 1/2 (1/2 sin(2x)-1/4 sin(4x))+C