Definite Integral of sin(arccosh(x)+7) How would I go about computing $\displaystyle\int_{10}^{16}\sin(\cosh^{-1}(x)+7)\mathrm dx$?
I haven't attempted anything yet, because I don't even know how to integrate the inverse hyperbolic cosine.

Making the problem of the antiderivative more general, let us consider
$$I=\int \sin \left(a+\cosh ^{-1}(x)\right)\,dx$$ As Fib1123 suggested, changing variable $$y=\cosh ^{-1}(x)\implies x=\cosh(y)\implies dx=\sinh(y)\,dy$$ the integrand becomes $$\sinh (y) \sin (a+y)=-i \sin(i y)\sin(a+y)$$ Now, using $$\sin(p)\sin(q)=\frac 12\left(\cos(p-q)-\cos(p+q)\right)$$ $$\sinh (y) \sin (a+y)=-\frac{i}{2}  (\cos (a+y-i y)-\cos (a+y +i y))$$ Integrate now to get $$I=\frac i 2 \left(\frac{\sin (a+y-i y)}{ (i-1)}+\frac{\sin (a+y+i y)}{ (i+1)}\right)$$ Now expand the sines taking into account $\sin(iy)=i \sinh(y)$ and $\cos(iy)=\cosh(y)$ and simplify the complex terms.
You should arrive to $$I=\frac{1}{2} (\cosh (y) \sin (a+y)-\sinh (y) \cos (a+y))$$
Edit
In the same spirit, the following integrals could be computed
$$\int \sin \left(a+b \cosh ^{-1}(x)\right)\,dx=\frac{\cosh (y) \sin (a+b y)-b \sinh (y) \cos (a+b y)}{b^2+1}$$ $$\int \cos \left(a+b \cosh ^{-1}(x)\right)\,dx=\frac{\cosh (y) \cos (a+b y)+b \sinh (y) \sin (a+b y)}{b^2+1}$$ $$\int \sin \left(a+b \sinh ^{-1}(x)\right)\,dx=\frac{\sinh (y) \sin (a+b y)-b \cosh (y) \cos (a+b y)}{b^2+1}$$ $$\int \cos \left(a+b \sinh ^{-1}(x)\right)\,dx=\frac{\sinh (y) \cos (a+b y)+b \cosh (y) \sin (a+b y)}{b^2+1}$$
