What is the integral of (sin(2x)*sinh(2x))/2 ?

The integral of (sin(2x)*sinh(2x))/2 is 1/8 (-cos(2x)sinh(2x)+sin(2x)cosh(2x))+C

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integral of (sin(2x)*sinh(2x))/2

\int \frac{sin ( 2 x ) \cdot sinh ( 2 x )}{2} d x

\frac{1}{8} (- cos (2 x) sinh (2 x) + sin (2 x) cosh (2 x)) + C

Solution steps

\int \frac{sin ( 2 x ) sinh ( 2 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 (2 x) sinh (2 x) d x

Apply Integration By Parts

: - \frac{1}{2} sinh (2 x) cos (2 x) - \int - cosh (2 x) cos (2 x) d x

= \frac{1}{2} (- \frac{1}{2} sinh (2 x) cos (2 x) - \int - cosh (2 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} (- \frac{1}{2} sinh (2 x) cos (2 x) - (- \int cosh (2 x) cos (2 x) d x))

Apply Integration By Parts

: \frac{1}{2} cosh (2 x) sin (2 x) - \int sinh (2 x) sin (2 x) d x

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

Therefore

\frac{1}{2} \cdot \int sinh (2 x) sin (2 x) d x = \frac{1}{2} (- \frac{1}{2} sinh (2 x) cos (2 x) - (- (\frac{1}{2} cosh (2 x) sin (2 x) - \int sinh (2 x) sin (2 x) d x)))

Isolate

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

= \frac{1}{2} (- \frac{cos ( 2 x ) sinh ( 2 x )}{4} + \frac{sin ( 2 x ) cosh ( 2 x )}{4})

Simplify

\frac{1}{2} (- \frac{cos ( 2 x ) sinh ( 2 x )}{4} + \frac{sin ( 2 x ) cosh ( 2 x )}{4}) : \frac{1}{8} (- cos (2 x) sinh (2 x) + sin (2 x) cosh (2 x))

= \frac{1}{8} (- cos (2 x) sinh (2 x) + sin (2 x) cosh (2 x))

Add a constant to the solution

= \frac{1}{8} (- cos (2 x) sinh (2 x) + sin (2 x) cosh (2 x)) + C

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What is the integral of (sin(2x)*sinh(2x))/2 ?
The integral of (sin(2x)*sinh(2x))/2 is 1/8 (-cos(2x)sinh(2x)+sin(2x)cosh(2x))+C