Indefinite integral off by a fraction

Question

The actual answer has a fraction of $\frac{1}{8}$ instead of $\frac{1}{5}$.

Can someone show me how my working out was incorrect?

Answer 1

It is false that$$\int\sin(u)e^u\,\mathrm du=\int\sin(4x)e^{4x}\,\mathrm dx.\tag1$$If you do $u=4x$, then you must also do $\mathrm du=4\,\mathrm dx$. So, instead of $(1)$, you actually have$$\int\sin(u)e^u\,\mathrm du=4\int\sin(4x)e^{4x}\,\mathrm dx.$$

Answer 2

Integration by complex numbers for the birth of the factor 1/8. $$ \begin{aligned} \int e^{4 x} \sin 4 x d x =& \operatorname{Im} \int e^{4 x} e^{4 i x} d x \\ =& \operatorname{Im} \int e^{4 x(1+i)} d x \\ =&\operatorname{lm}\left(\frac{e^{4x(1+i)}}{4 (1+i)}\right)+C \\ =& \operatorname{lm}\left(\frac{e^{4 x} \cdot e^{4 x i}}{4(1+i)}\right)+C\\ =& \operatorname{lm}\left(\frac{e^{4 x}(\cos 4 x+i \sin 4 x)}{4(1+i)} \cdot \frac{(1-i)}{1-i}\right)+C\\ =& \frac{e^{4 x}}{8}(\sin 4 x-\cos 4 x)+C \end{aligned} $$