What is the integral of sin(x)sin(4x) ?

The integral of sin(x)sin(4x) is 1/2 (1/3 sin(3x)-1/5 sin(5x))+C

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integral of sin(x)sin(4x)

\int sin (x) sin (4 x) d x

\frac{1}{2} (\frac{1}{3} sin (3 x) - \frac{1}{5} sin (5 x)) + C

Solution steps

\int sin (x) sin (4 x) d x

Rewrite using trig identities

= \int \frac{1}{2} (- cos (5 x) + cos (- 3 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 (- 3 x) d x

Simplify

- cos (5 x) + cos (- 3 x) : cos (3 x) - cos (5 x)

= \frac{1}{2} \cdot \int cos (3 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 (3 x) d x - \int cos (5 x) d x)

\int cos (3 x) d x = \frac{1}{3} sin (3 x)

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

= \frac{1}{2} (\frac{1}{3} sin (3 x) - \frac{1}{5} sin (5 x))

Add a constant to the solution

= \frac{1}{2} (\frac{1}{3} sin (3 x) - \frac{1}{5} sin (5 x)) + C

\frac{d}{d x} (e^{x + 7})

\frac{\partial}{\partial x} (- \frac{e ^{\frac{y}{x}} y}{x ^{2}} + e^{- x} y)

n = 0 \sum \infty e^{\frac{1}{n}}

2 e^{- 2 t} cos (3 t)

\int \frac{1}{2} e^{x} sin (x) d x

What is the integral of sin(x)sin(4x) ?
The integral of sin(x)sin(4x) is 1/2 (1/3 sin(3x)-1/5 sin(5x))+C