Answer: (a) \(\sinh 2x=2\sinh x\cosh x\).
(b) \(\displaystyle \int \sinh^2 2x\cosh x\,dx=4\left(\frac15\sinh^5 x+\frac13\sinh^3 x\right)+C\).
(c) \(\displaystyle y=4\operatorname{sech}x\left(\frac15\sinh^5 x+\frac13\sinh^3 x+1\right)\).
(a) Using the definitions
\(\cosh x=\frac12(e^x+e^{-x})\) and \(\sinh x=\frac12(e^x-e^{-x})\).
Then
\(2\sinh x\cosh x=2\cdot \frac12(e^x-e^{-x})\cdot \frac12(e^x+e^{-x})\)
\(=\frac12(e^x-e^{-x})(e^x+e^{-x})\)
\(=\frac12(e^{2x}-e^{-2x})\)
\(=\sinh 2x\).
So \(\sinh 2x=2\sinh x\cosh x\).
(b) Let \(u=\sinh x\). Then \(du=\cosh x\,dx\).
From part (a), \(\sinh 2x=2\sinh x\cosh x\), so
\(\sinh^2 2x=4\sinh^2 x\cosh^2 x\).
Hence
\(\displaystyle \int \sinh^2 2x\cosh x\,dx=4\int \sinh^2 x\cosh^2 x\,du\).
Now use \(\cosh^2 x=1+\sinh^2 x\):
\(4\int \sinh^2 x(1+\sinh^2 x)\,du=4\int u^2(u^2+1)\,du\).
So
\(4\int (u^4+u^2)\,du=4\left(\frac15u^5+\frac13u^3\right)+C\).
Substituting back \(u=\sinh x\),
\(\displaystyle \int \sinh^2 2x\cosh x\,dx=4\left(\frac15\sinh^5 x+\frac13\sinh^3 x\right)+C\).
(c) The differential equation is
\(\displaystyle \frac{dy}{dx}+y\tanh x=\sinh^2 2x\).
This is linear, with integrating factor
\(\displaystyle e^{\int \tanh x\,dx}=e^{\ln(\cosh x)}=\cosh x\).
Multiplying through by \(\cosh x\),
\(\cosh x\frac{dy}{dx}+y\sinh x=\sinh^2 2x\cosh x\).
The left-hand side is
\(\displaystyle \frac{d}{dx}(y\cosh x)\), so
\(\displaystyle \frac{d}{dx}(y\cosh x)=\sinh^2 2x\cosh x\).
Integrating,
\(\displaystyle y\cosh x=4\left(\frac15\sinh^5 x+\frac13\sinh^3 x\right)+C\).
Use \(y=4\) when \(x=0\). Since \(\sinh 0=0\) and \(\cosh 0=1\),
\(4=C\).
Therefore
\(\displaystyle y\cosh x=4\left(\frac15\sinh^5 x+\frac13\sinh^3 x+1\right)\),
so
\(\displaystyle y=4\operatorname{sech}x\left(\frac15\sinh^5 x+\frac13\sinh^3 x+1\right)\).
