Integrate twice and solve for the repeated integral · source reference

Printed question

Find $\int e^{2x}\cos3x dx$. [5]

[2023/EJC/Prelim/I/5b]

Printed working

$$\begin{aligned}I=\int e^{2x}\cos3x dx&=\cos3x\frac{e^{2x}}2-\int(-3\sin3x)\frac{e^{2x}}2dx\\&=\frac12e^{2x}\cos3x+\frac32\int\sin3x e^{2x}dx\\&=\frac12e^{2x}\cos3x+\frac32\left[\frac{e^{2x}}2\sin3x-\int(3\cos3x)\frac{e^{2x}}2dx\right]\\&=\frac12e^{2x}\cos3x+\frac34e^{2x}\sin3x-\frac94\int\cos3x e^{2x}dx.\end{aligned}$$

Moving the integral over:

$$\frac{13}4I=\frac12e^{2x}\cos3x+\frac34e^{2x}\sin3x+c.$$ $$I=\frac4{13}\left(\frac12e^{2x}\cos3x+\frac34e^{2x}\sin3x\right)+c_1=\frac1{13}e^{2x}(2\cos3x+3\sin3x)+c_1.$$

Teacher notes

Integration by parts. For the first integral, $u=\cos3x$, $dv/dx=e^{2x}$. For the second, $u=\sin3x$, $dv/dx=e^{2x}$. Since the LHS integral is indefinite, put “$+c$” on the RHS expression. Note $c_1=(4/13)c$.

Printed answer key

The question-file short key unexpectedly lists two parts:

(a) $-\frac18(\cos4x+2\cos2x)+c$;

(b) $e^{2x}(2\cos3x+3\sin3x)+c_1$.

There is no part (a) in the printed question. Part (b) omits the factor $1/13$ shown in the detailed solution.

Mathematical review

The detailed solution is correct. Differentiating the short-key expression in (b) gives 13 times the integrand. The unmatched (a) is retained as an answer-key anomaly, not invented into a question.