Integration by Parts — Question 5

Question 5 – Solution

Let $$I=\int e^{2x}\cos(3x)\,dx.$$ For the first integration by parts, choose $$u=\cos(3x),\quad dv=e^{2x}\,dx, \qquad du=-3\sin(3x)\,dx,\quad v=\frac12e^{2x}.$$ Using $\int u\,dv=uv-\int v\,du$, we get $$\begin{align*} I &=\frac12e^{2x}\cos(3x) -\int \frac12e^{2x}\bigl(-3\sin(3x)\bigr)\,dx \\ &=\frac12e^{2x}\cos(3x) +\frac32\int e^{2x}\sin(3x)\,dx. \tag{1} \end{align*}$$

Now evaluate the remaining integral. Let $$J=\int e^{2x}\sin(3x)\,dx.$$ Use integration by parts again with $$u=\sin(3x),\quad dv=e^{2x}\,dx, \qquad du=3\cos(3x)\,dx,\quad v=\frac12e^{2x}.$$ Then $$\begin{align*} J &=\frac12e^{2x}\sin(3x) -\frac32\int e^{2x}\cos(3x)\,dx \\ &=\frac12e^{2x}\sin(3x)-\frac32I. \tag{2} \end{align*}$$

Substitute equation (2) into equation (1): $$\begin{align*} I &=\frac12e^{2x}\cos(3x) +\frac32\left(\frac12e^{2x}\sin(3x)-\frac32I\right) \\ &=\frac12e^{2x}\cos(3x) +\frac34e^{2x}\sin(3x)-\frac94I. \end{align*}$$ Move the $I$-term to the left and solve: $$\begin{align*} I+\frac94I &=\frac12e^{2x}\cos(3x)+\frac34e^{2x}\sin(3x), \\ \frac{13}{4}I &=\frac12e^{2x}\cos(3x)+\frac34e^{2x}\sin(3x), \\ I &=\frac{e^{2x}}{13}\bigl(2\cos(3x)+3\sin(3x)\bigr). \end{align*}$$ Therefore, $$\boxed{\displaystyle \int e^{2x}\cos(3x)\,dx =\frac{e^{2x}}{13}\bigl(2\cos(3x)+3\sin(3x)\bigr)+C}$$

To check, differentiate the antiderivative: $$\begin{align*} \frac{d}{dx}\left[ \frac{e^{2x}}{13}\bigl(2\cos(3x)+3\sin(3x)\bigr)\right] &=\frac{e^{2x}}{13} \bigl[2(2\cos(3x)+3\sin(3x))\\ &\qquad\quad-6\sin(3x)+9\cos(3x)\bigr] \\ &=e^{2x}\cos(3x), \end{align*}$$ which is the original integrand.