Finding the antiderivative of the product of two functions given only their derivative properties

Question

let $\alpha'(x)=\beta(x), \beta'(x)=\alpha(x)$ and assume that $\alpha^2 - \beta^2 = 1$. how would I go about calculating the following anti derivative : $\int (\alpha (x))^5 (\beta(x))^4$d$x$.

Thank you.

Answer 1

$$\alpha''(x)=\beta'(x)=\alpha(x)$$ The general solution of the ODE : $\quad\alpha''(x)=\alpha(x)\quad$ is : $$\quad\alpha(x)=c_1\cosh(x)+c_2\sinh(x)$$ And then : $$\quad\beta(x)=\alpha'(x)=c_1\sinh(x)+c_2\cosh(x)$$ This, combined with $\quad \alpha^2-\beta^2=1\quad$ implies : $\begin{cases} \alpha(x)=\cosh(x) \\ \beta(x)=\sinh(x) \\ \end{cases}$

$$\int \alpha^5\beta^4 dx=\int \cosh^5\sinh^4 dx=\int \left(1+\sinh^2(x) \right)^2\sinh^4(x) \cosh(x)dx$$ Continue with change of variable $\sinh(x)=t$