Category Archives: Integration by Parts

October 28, 2025 · 5:42 am

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)

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

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

Integrating the Natural Log Function

One of the students asked me the other day how to integrate $f(x)=ln(x)$ – it’s not part of their course, but I thought I would do it here.

We use integration by parts to integrate $ln(x)$

$\begin{equation*}\int u dv=uv-\int v du\end{equation}$

$\begin{equation*}\int ln(x) dx\end{equation}$

Let $u=ln(x)$ and $dv=1$ , then $du=\frac{1}{x}$ and $v=x$

$\begin{equation*}\int ln(x) dx = xln(x)-\int x\times \frac{dx}{x}\end{equation}$

$\begin{equation*}\int ln(x) dx = xln(x)-\int 1 dx\end{equation}$

$\begin{equation*}\int ln(x) dx = xln(x)-x+c\end{equation}$

What about $\int 5ln(\sqrt{x}) dx$

We can take advantage of log laws and the properties of integration.

$\begin{equation*}\int 5ln(\sqrt{x}) dx=5\int lnx^{\frac{1}{2}} dx=5\int \frac{1}{2}ln(x) dx=\frac{5}{2} \int ln(x) dx\end{equation}$

$\begin{equation*}\frac{5}{2} \int ln(x) dx=\frac{5}{2}(xlnx-x)+c\end{equation}$

More Integration

I went down a rabbit hole while reading An Imaginary Tale by Paul J Nahin and I decided I wanted to do this…

$\begin{equation*}\int_0^1{x^x dx}\end{equation}$

$\begin{equation*}x^x=e^{ln(x^x)}\end{equation}$

$\begin{equation*}x^x=e^{xln(x)}\end{equation}$

The power series expansion of $e^x$ is

$\begin{equation*}e^x=1+x+\frac{1}{2!}x^2+\frac{1}{3!}x^3+\frac{1}{4!}x^4+...\end{equation}$

$\begin{equation*}\therefore e^{xln(x)}=1+xln(x)+\frac{1}{2!}(xln(x))^2+\frac{1}{3!}(xln(x))^3+...\end{equation}$

$\begin{equation*}\therefore e^{xln(x)}=\Sigma_{n=0}^{\infty}(\frac{1}{n!}(xln(x))^n)\end{equation}$

Hence $\int_0^1{x^x dx}=\int_0^1(\Sigma_{n=0}^{\infty}(\frac{1}{n!}(xln(x))^n dx)$

$\begin{equation*}=\Sigma_{n=0}^{\infty}(\frac{1}{n!}\int_0^1xln(x))^n dx)\end{equation}$

Let’s consider the integral

(1) $\begin{equation*}\int_0^1 x^n(ln(x))^n dx\end{equation*}$

Let $u=ln(x)$ then $\frac{du}{dx}=\frac{1}{x}$ and $dx=x du$ where $x=e^u$

When $x=0, u=-\infty$ and when $x=1, u=0$

(2) $\begin{equation*}\int_0^1 x^n(ln(x))^n dx=\int_{-\infty}^0 e^{nu}u^ne^u du\end{equation*}$

(3) $\begin{equation*}\int_0^1 x^n(ln(x))^n dx=\int_{-\infty}^0 e^{(n+1)u}u^n du\end{equation*}$

Integrate by parts using the tabular method.

Sign	Differentiate	Integrate
+	$u^n$	$e^{(n+1)u}$
–	$nu^{n-1}$	$\frac{e^{(n+1)u}}{n+1}$
+	$n(n-1)u^{n-2}$	$\frac{e^{(n+1)u}}{(n+1)^2}$
–	$n(n-1)(n-2)u^{n-3}$	$\frac{e^{(n+1)u}}{(n+1)^3}$
+	$\frac{n!}{(n-4)!}u^{n-4}$	$\frac{e^{(n+1)u}}{(n+1)^4}$
	$\vdots$	$\vdots$
	$\frac{n!}{(n-n)!}u^{n-n}$	$\frac{e^{(n+1)u}}{(n+1)^{n}}$
$(-1)^n$	$0$	$\frac{e^{(n+1)u}}{(n+1)^{n+1}}$

When we substitute $u=-\infty$ or $u=0$ the differentiation column is zero except for $\frac{n!}{(n-n)!}u^{n-n}$ , which is $n!$ ,

Thus $\int_0^1 x^n(ln(x))^n dx=\frac{e^{(n+1)u}}{(n+1)^{n+1}}}]_{-\infty}^0$

$\begin{equation*}=n!\times\frac{e^0}{(n+1)^{n+1}}-0\end{equation}$

$\begin{equation*}=\frac{n!}{(n+1)^{n+1}}\end{equation}$

Now we just need to think about the sign.

$\begin{equation*}=(-1)^n\frac{n!}{(n+1)^{n+1}}\end{equation}$

The integral is now

$\int_0^1{x^x dx}=(\Sigma_{n=0}^{\infty}(\frac{1}{n!}( (-1)^n\frac{n!}{(n+1)^{n+1}})$

So $\int_0^1{x^x dx}=\Sigma_{n=0}^{\infty}( (-1)^n\frac{1}{(n+1)^{n+1}}$

Let’s work out some partial sums

$\int_0^1{x^x dx}=0.778343$

Integration by Parts using the Tabular Method

(1) $\begin{equation*}\int_0^1x^2e^xdx\end{equation*}$

I am going to do this integral in two ways; the traditional method and the tabular method.

Traditional Method

Remember $\int{u dv}=u\times v-\int{v du}$

Let $u=x^2$ and $dv=e^x$

Then $du=2x$ and $v=\int{e^x dx}=e^x$

$\begin{equation*}\int_0^1{x^2e^x dx}=x^2 \times e^x ]_0^1-\int_0^1{e^x 2x dx}\end{equation}$

Now we need to do integration by parts on $\int_0^1{e^x 2x dx}$

Let $u=2x$ and $dv=e^x$

Then $du=2$ and $v=e^x$

$\begin{equation*}\int_0^1{x^2e^x dx}=x^2 \times e^x ]_0^1-(2x\times e^x]_0^1-\int_0^1{e^x 2 dx})\end{equation}$

$\begin{equation*}\int_0^1{x^2e^x dx}=x^2 \times e^x ]_0^1-(2x\times e^x]_0^1-(2e^x )]_0^1)\end{equation}$

$\begin{equation*}\int_0^1{x^2e^x dx}=e-(2e-(2e-2))\end{equation}$

$\begin{equation*}\int_0^1{x^2e^x dx}=e+2\end{equation}$

Tabular Integration

Similar to before, select a $u$ and a $dv$ , $u=x^2$ and $dv=e^x$

Sign	D(ifferentiate)	I(ntegrate)
+	$x^2$	$e^x$
–	$2x$	$e^x$
+	$2$	$e^x$
–	$0$	$e^x$

Stop when the differentiating column reaches zero.

Then we multiply diagonally

$(+x^2)(e^x)+(-2x)(e^x)+(+2e^x)$

$=x^2e^x-2xe^x+2e^x]_0^1$

$=e-2e+2e-(0-0-2)$

$=e+2$

It is only worth using this method if integration by parts is required more than once. Also, the $u$ has to eventually differentiate to $0$ .

Let’s try another one

(2) $\begin{equation*}\int{x^3cos(2x) dx}\end{equation*}$

Let $u=x^3$ and $dv=cos(2x)$

Sign	D	I
+	$x^3$	$cos(2x)$
–	$3x^2$	$\frac{1}{2}sin(2x)$
+	$6x$	$-\frac{1}{4}cos(2x)$
–	$6$	$-\frac{1}{8}sin(2x)$
+	$0$	$\frac{1}{16}cos(2x)$

$\int{x^3cos(2x) dx}=(x^3)(\frac{1}{2}sin(2x))+(-3x^2)(-\frac{1}{4}cos(2x))+(6x)(-\frac{1}{8}sin(2x))+(-6)(\frac{1}{16}cos(2x))+c$

$=\frac{x^3}{2}sin(2x)+\frac{3x^2}{4}cos(2x)-\frac{3x}{4}sin(2x)-\frac{3}{8}cos(2x)+c$

1 Comment

Filed under Integration, Integration by Parts, Tabular Integration

Tagged as Calculus, integration by parts, tabular integration

$n$	$((-1)^n\frac{1}{(n+1)^{n+1}})$
$5$	$=0.78343$
$10$	$=0.78343$
$20$	$=0.78343$
$100$	$=0.78343$

Category Archives: Integration by Parts

Integration Question (much easier with Integration by Parts)

Integration by Parts

Integrating the Natural Log Function

More Integration

Integration by Parts using the Tabular Method

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