Solve $\int_{0}^{1} t^2 \sqrt{1+4t^2} dt$
Making the substitution $t=\frac{\sinh(x)}{2},$ then $4t^2=\sinh^2(x),$ and we get $dt=\frac{\cosh(x)}{2}dx$
$$\begin{align} \int t^2 \sqrt{(1+4t^2)}dt&=\frac{1}{8}\int{\sinh^2(x)}\sqrt{1+\sinh^2(x)}\cosh(x)dx\\ &=\frac{1}{8}\int{\sinh^2(x)}\cosh^2(x)dx=\frac{1}{8}\left(\frac{1}{32}\sinh(4x)-4x\right) +C \\ &= \frac{1}{8}\left(\frac{1}{32}(4\sinh(x)\cosh(x) - 8\sinh^3(x)\cosh(x)) - 4x\right) + C\\ &=\left[\frac{1}{8}\left(\frac{1}{32}(4(2t)\sqrt{1+4t^2} - 8(2t)^3\sqrt{1+4t^2}) - 4\sinh^{-1}(2t)\right)\right]_{0}^{1} \approx -1.222 \end{align}$$ Answer is like 0.6 not what I have.. where did I go wrong?
There are 2 mistakes.
$$\begin{align} \int t^2 \sqrt{(1+4t^2)}dt&=\frac{1}{8}\int{\sinh^2(x)}\sqrt{1+\sinh^2(x)}\cosh(x)dx\\ &=\frac{1}{8}\int{\sinh^2(x)}\cosh^2(x)dx=\frac{1}{8\times 32}\left(\sinh(4x)-4x\right) +C \\ &= \frac{1}{256}\left((4\sinh(x)\cosh^3(x) + 4\sinh^3(x)\cosh(x)) - 4x\right) + C\\ &=\left[\frac{1}{256}\left(4(2t)(\sqrt{1+4t^2})^3 + 4(2t)^3\sqrt{1+4t^2} - 4\sinh^{-1}(2t)\right)\right]_{0}^{1} \approx 0.6063 \end{align}$$