What is the integral of sin(10x)cos(5x) ?

The integral of sin(10x)cos(5x) is 1/2 (-1/15 cos(15x)-1/5 cos(5x))+C

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integral of sin(10x)cos(5x)

\int sin (10 x) cos (5 x) d x

\frac{1}{2} (- \frac{1}{15} cos (15 x) - \frac{1}{5} cos (5 x)) + C

Solution steps

\int sin (10 x) cos (5 x) d x

Use the following identity:

cos (t) sin (s) = \frac{sin ( s + t ) + sin ( s - t )}{2}

= \int \frac{sin ( 10 x + 5 x ) + sin ( 10 x - 5 x )}{2} 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 sin (10 x + 5 x) + sin (10 x - 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 sin (10 x + 5 x) d x + \int sin (10 x - 5 x) d x)

\int sin (10 x + 5 x) d x = - \frac{1}{15} cos (15 x)

\int sin (10 x - 5 x) d x = - \frac{1}{5} cos (5 x)

= \frac{1}{2} (- \frac{1}{15} cos (15 x) - \frac{1}{5} cos (5 x))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{15} cos (15 x) - \frac{1}{5} cos (5 x)) + C

\frac{\partial}{\partial y} (x^{4} - 3 x^{2} y^{2} + y^{4})

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

\frac{d}{d x} (\frac{sin ( x )}{2})

y = \frac{4 x}{x ^{2} + 1}

f (x) = x^{2.4} + e^{2.4}

What is the integral of sin(10x)cos(5x) ?
The integral of sin(10x)cos(5x) is 1/2 (-1/15 cos(15x)-1/5 cos(5x))+C