Integration Question (much easier with Integration by Parts)

The Year 12 Mathematics Methods course doesn’t cover Integration by Parts, so they end up with questions like the following.

Determine the following:
(a) $\frac{d}{dx} \left (e^{2x}sin(3x) \right )$
(b) $\frac{d}{dx} \left (e^{2x}cos(3x) \right )$
Hence, determine the following integral by considering both parts (a) and (b)
$\int_0^{\frac{\pi}{2}}13e^2xcos(3x) \enspace dx$

(a) Use the product rule

(1) $\begin{equation*}\frac{d}{dx} \left (e^{2x}sin(3x) \right)= 2e^{2x}sin(3x)+3e^{2x}cos(3x) \end{equation*}$

(b)

(2) $\begin{equation*}\frac{d}{dx} \left (e^{2x}cos(3x) \right )=2e^{2x}cos(3x)-3e^{2x}sin(3x)\end{equation*}$

I need to use equations $1$ and $2$ to find $\int_0^{\frac{\pi}{2}}13e^2xcos(3x) \enspace dx$ .

The $e^{2x}sin(3x)$ terms need to vanish and I need $13$ of the $e^{2x}cos(3x)$ terms.

$3\times \text{equation} (1)+ 2\times \text{equation} (2)$

(3) $\begin{equation*}3\frac{d}{dx} \left (e^{2x}sin(3x) \right)= 6e^{2x}sin(3x)+9e^{2x}cos(3x) \end{equation*}$

(4) $\begin{equation*}2\frac{d}{dx} \left (e^{2x}cos(3x) \right )=4e^{2x}cos(3x)-6e^{2x}sin(3x)\end{equation*}$

Equation $3$ plus equation $4$

(5) $\begin{equation*}3\frac{d}{dx} \left (e^{2x}sin(3x) \right)+2\frac{d}{dx} \left (e^{2x}cos(3x) \right )=13e^{2x}cos(3x)\end{equation*}$

Integrate both sides of the equation

$\int_0^{\frac{\pi}{2}} \left (3\frac{d}{dx} \left (e^{2x}sin(3x) \right)+2\frac{d}{dx} \left (e^{2x}cos(3x) \right ) \right ) dx=\int_0^{\frac{\pi}{2}}13e^{2x}cos(3x) \enspace dx$

By the fundamental theorem of calculus, we know

$(3e^{2x}sin(3x) \right)+2 \left (e^{2x}cos(3x) \right ]_0^{\frac{\pi}{2}}=\int_0^{\frac{\pi}{2}}13e^{2x}cos(3x) \enspace dx$

$3e^{\pi}sin(\frac{3\pi}{2}})+2e^{\pi}cos(\frac{3\pi}{2}})-3e^0sin(3(0))-2e^0cos(3(0))=\int_0^{\frac{\pi}{2}}13e^{2x}cos(3x) \enspace dx$

$\int_0^{\frac{\pi}{2}}13e^{2x}cos(3x) \enspace dx=-3e^{\pi}-2$

Integration by Parts

Remember $\int u\enspace dv=uv-\int v \enspace du$

$\int_0^{\frac{\pi}{2}}e^{2x}cos(3x) \enspace dx$

Let $u=cos(3x)$ , then $du=-3sin(3x)$

and $dv=e^{2x}$ , then $v=\frac{e^{2x}}{2}$

$\int_0^{\frac{\pi}{2}}e^{2x}cos(3x) \enspace dx = \left (cos(3x)\left (\frac{e^{2x}}{2}\right ) \right )_0^{\frac{\pi}{2}}-\int_0^{\frac{\pi}{2}} \frac{e^{2x}}{2}(-3sin(3x)) \enspace dx$

$\int_0^{\frac{\pi}{2}}e^{2x}cos(3x) \enspace dx=cos(\frac{3\pi}{2})(\frac{e^\pi}{2})-cos(0)(\frac{e^0}{2})+\frac{3}{2}\int_0^{\frac{\pi}{2}}sin(3x)e^{2x} \enspace dx$

$\int_0^{\frac{\pi}{2}}e^{2x}cos(3x) \enspace dx=-\frac{1}{2}+\frac{3}{2}\int_0^{\frac{\pi}{2}}sin(3x)e^{2x} \enspace dx$

Let $u=sin(3x)$ , then $du=3cos(3x)$

and $dv=e^{2x}$ . then $v=\frac{e^{2x}}{2}$

$\int_0^{\frac{\pi}{2}}e^{2x}cos(3x) \enspace dx=-\frac{1}{2}+\frac{3}{2}(\frac{e^{2x}}{2}sin(3x)]_0^{\frac{\pi}{2}}-\int_0^{\frac{\pi}{2}} \frac{e^{2x}}{2}(3cos(3x)) \enspace dx)$

$\int_0^{\frac{\pi}{2}}e^{2x}cos(3x) \enspace dx=-\frac{1}{2}+\frac{3}{2}(-\frac{e^\pi}{2}-\frac{3}{2} \int_0^{\frac{\pi}{2}} e^{2x}cos(3x) \enspace dx)$

$\int_0^{\frac{\pi}{2}}e^{2x}cos(3x) \enspace dx=-\frac{1}{2}-\frac{3e^\pi}{4}-\frac{9}{4} \int_0^{\frac{\pi}{2}} e^{2x}cos(3x) \enspace dx$

Collect like terms (the integrals are like)

$\frac{13}{4}(\int_0^{\frac{\pi}{2}}e^{2x}cos(3x) \enspace dx)=-\frac{1}{2}-\frac{3e^\pi}{4}$

$(\int_0^{\frac{\pi}{2}}e^{2x}cos(3x) \enspace dx)=\frac{-2-3e^\pi}{13}$

Integration Question (much easier with Integration by Parts)

Integration by Parts

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