Question

$$\int \sin ax \sinh bx \ dx $$

Accepted Answer

$$\frac{1}{a^2 + b^2} \left[b \cosh bx \sin ax -a \cos ax \sinh bx\right]$$

Front	\[\int \sin ax \sinh bx \ dx \]
Back	\[\frac{1}{a^2 + b^2} \left[b \cosh bx \sin ax -a \cos ax \sinh bx\right]\]

Ax Sin Sinh Bx Int Dx Frac{1}{A^2 B^2