Learning Outcomes
- Compute integrals using integration by parts.
Section 5.2 Integration by Parts (TI2)
Subsection 5.2.1 Activities
Activity 5.2.1.
Answer the following.
(a)
Using the product rule, which of these is derivative of \(x^3e^x\) with respect to \(x\text{?}\)
- \(\displaystyle 3x^2e^x\)
- \(\displaystyle 3x^2e^{x}+x^3e^x\)
- \(\displaystyle 3x^2e^{x-1}\)
- \(\displaystyle \frac{1}{4}x^4 e^x\)
(b)
Based on this result, which of these would you suspect to equal \(\int 3x^2e^x+x^3e^x\,dx\text{?}\)
- \(\displaystyle x^3e^x+C\)
- \(\displaystyle x^3e^x+\frac{1}{4}x^4e^x+C\)
- \(\displaystyle 6xe^x+3x^2e^x+C\)
- \(\displaystyle 6xe^x+3x^2e^x+3x^2e^x+x^3e^x+C\)
Activity 5.2.2.
Answer the following.
(a)
Which differentiation rule is easier to implement?
- Product Rule
- Chain Rule
(b)
Which differentiation strategy do expect to be easier to reverse?
- Product Rule
- Chain Rule
Activity 5.2.3.
Answer the following.
(a)
Which of the following equations is equivalent to the formula \(\frac{d}{dx}[uv]=u'v+uv'\text{?}\)
- \(\displaystyle uv'=-\frac{d}{dx}(uv)-vu'\)
- \(\displaystyle uv'=-\frac{d}{dx}(uv)+vu'\)
- \(\displaystyle uv'=\frac{d}{dx}(uv)+vu'\)
- \(\displaystyle uv'=\frac{d}{dx}(uv)-vu'\)
(b)
Which of these is the most concise result of integrating both sides with respect to \(x\text{?}\)
- \(\displaystyle \int(uv')\,dx=uv-\int(vu')\,dx\)
- \(\displaystyle \int(u)\,dv=uv-\int(v)\,du\)
- \(\displaystyle \int(uv')\,dx=uv-\int(vu')\,dx+C\)
- \(\displaystyle \int(u)\,dv=uv-\int(v)\,du+C\)
Fact 5.2.4.
By the product rule, \(\frac{d}{dx}[uv]=u'v+uv'\) and, subsequently, \(uv'=\frac{d}{dx}[uv]-u'v\text{.}\) There is a dual integration technique reversing this process, known as integration by parts.
This technique involves using algebra to rewrite an integral of a product of functions in the form \(\int (u)\,dv\) and then using the equality
\begin{equation*}
\int (u)\,dv=uv-\int(v)\,du\text{.}
\end{equation*}
Activity 5.2.5.
Consider \(\int xe^{x}\,dx\text{.}\) Suppose we decided to let \(u=x\text{.}\)
(a)
Compute \(\frac{du}{dx}=\unknown\text{,}\) and rewrite it as \(du=\unknown\,dx\text{.}\)
(b)
What is the best candidate for \(dv\text{?}\)
- \(\displaystyle dv=x\,dx\)
- \(\displaystyle dv=e^x\)
- \(\displaystyle dv=x\)
- \(\displaystyle dv=e^x\,dx\)
(c)
Given that \(dv=e^x\,dx\text{,}\) find \(v=\unknown\text{.}\)
(d)
Show why \(\int xe^{x}\,dx\) may now be rewritten as \(xe^x-\int e^x\,dx\text{.}\)
(e)
Solve \(\int e^x\,dx\text{,}\) and then give the most general antiderivative of \(\int xe^{x}\,dx\text{.}\)
Example 5.2.6.
Here is how one might write out the explanation of how to find \(\int xe^{x}\,dx\) from start to finish:
\begin{align*}
\int xe^{x}\,dx\\
& u=x & dv = e^x\,dx\\
& du=1\cdot\,dx & v=e^x\\
\int xe^x \,dx &= xe^x-\int e^x \,dx\\
&= xe^{x}-e^x+C
\end{align*}
Activity 5.2.7.
Which step of the previous example do you think was the most important?
- Choosing \(u=x\) and \(dv=e^x\,dx\text{.}\)
- Finding \(du=1\,dx\) and \(v=e^x\,dx\text{.}\)
- Applying integration by parts to rewrite \(\int xe^x\,dx\) as \(xe^x-\int e^x\,dx\text{.}\)
- Integrating \(\int e^x\,dx\) to get \(xe^{x}-e^x+C\text{.}\)
Activity 5.2.8.
Consider the integral \(\int x^9\ln(x) \,dx\text{.}\) Suppose we proceed using integration by parts. We choose \(u=\ln(x)\) and \(dv=x^9\,dx\text{.}\)
(a)
What is \(du\text{?}\)
(b)
What is \(v\text{?}\)
(c)
What do you get when plugging these pieces into integration by parts?
(d)
Does the new integral \(\int v\,du\) seem easier or harder to compute than the original integral \(\int x^9\ln(x) \,dx\text{?}\)
- The original integral is easier to compute.
- The new integral is easier to compute.
- Neither integral seems harder than the other one.
Activity 5.2.9.
Consider the integral \(\int x^9\ln(x) \,dx\) once more. Suppose we still proceed using integration by parts. However, this time we choose \(u=x^9\) and \(dv=\ln(x)\,dx\text{.}\) Do you prefer this choice or the choice we made in Activity 5.2.8?
- We prefer the substitution choice of \(u=\ln(x)\) and \(dv=x^9\,dx\text{.}\)
- We prefer the substitution choice of \(u=x^9\) and \(dv=\ln(x)\,dx\text{.}\)
- We do not have a strong preference, since these choices are of the same difficulty.
Activity 5.2.10.
Consider the integral \(\int x\cos(x)\,dx\text{.}\) Suppose we proceed using integration by parts. Which of the following candidates for \(u\) and \(dv\) would best allow you to evaluate this integral?
- \(u=\cos(x)\text{,}\) \(dv=xdx\)
- \(u=\cos(x)\,dx\text{,}\) \(dv=x\)
- \(u=x\,dx\text{,}\) \(dv=\cos(x)\)
- \(u=x\text{,}\) \(dv=\cos(x)\,dx\)
Activity 5.2.11.
Evaluate the integral \(\int x\cos(x)\,dx\) using integration by parts.
Activity 5.2.12.
Now use integration by parts to evaluate the integral \(\displaystyle \int_{\frac{\pi}{6}}^{\pi} x\cos(x)\,dx\text{.}\)
Activity 5.2.13.
Consider the integral \(\int x\arctan(x)\,dx\text{.}\) Suppose we proceed using integration by parts. Which of the following candidates for \(u\) and \(dv\) would best allow you to evaluate this integral?
- \(u=x\,dx\text{,}\) \(dv=\arctan(x)\)
- \(u=\arctan(x)\text{,}\) \(dv=x\,dx\)
- \(u=x\arctan(x)\text{,}\) \(dv=\,dx\)
- \(u=x\text{,}\) \(dv=\arctan(x)\,dx\)
Activity 5.2.14.
Consider the integral \(\int e^x\cos(x)\,dx\text{.}\) Suppose we proceed using integration by parts. Which of the following candidates for \(u\) and \(dv\) would best allow you to evaluate this integral?
- \(u=e^x\text{,}\) \(dv=\cos(x)\,dx\)
- \(u=\cos(x)\text{,}\) \(dv=e^x\,dx\)
- \(u=e^x\,dx\text{,}\) \(dv=\cos(x)\)
- \(u=\cos(x)\,dx\text{,}\) \(dv=e^x\)
Activity 5.2.15.
Suppose we started using integration by parts to solve the integral \(\int e^x\cos(x)\,dx\) as follows:
\begin{align*}
\int e^x\cos(x)\,dx\\
& u=\cos(x) & dv = e^x\,dx\\
& du=-sin(x) \,dx & v=e^x\\
\int e^x\cos(x)\,dx &= \cos(x)e^x-\int e^x(-\sin(x) \,dx)\\
&= \cos(x)e^x+\int e^x\sin(x) \,dx
\end{align*}
We will have to use integration by parts a second time to evaluate the integral \(\int e^x\sin(x) \,dx\text{.}\) Which of the following candidates for \(u\) and \(dv\) would best allow you to continue evaluating the original integral \(\int e^x\cos(x)\,dx\text{?}\)
- \(u=e^x\text{,}\) \(dv=\sin(x)\,dx\)
- \(u=\sin(x)\text{,}\) \(dv=e^x\,dx\)
- \(u=e^x\,dx\text{,}\) \(dv=\sin(x)\)
- \(u=\sin(x)\,dx\text{,}\) \(dv=e^x\)
Activity 5.2.16.
Use integration by parts to show that \(\displaystyle \int_0^{\frac{\pi}{4}} x\sin(2x)\,dx=\frac{1}{4}\text{.}\)
Activity 5.2.17.
Consider the integral \(\int t^5 \sin(t^3)\,dt\text{.}\)
(a)
Use the substitution \(x=t^3\) to rewrite the integral in terms of \(x\text{.}\)
(b)
Use integration by parts to evaluate the integral in terms of \(x\text{.}\)
(c)
Replace \(x\) with \(t^3\) to finish evaluating the original integral.
Activity 5.2.18.
Use integration by parts to show that \(\displaystyle \int \ln(z)\,dz=z \ln(z) - z + C\text{.}\)
Activity 5.2.19.
Given that that \(\displaystyle \int \ln(z)\,dz=z \ln(z) - z + C\text{,}\) evaluate \(\displaystyle \int (\ln(z))^2\,dz\text{.}\)
Activity 5.2.20.
Consider the antiderivative \(\displaystyle\int (\sin(x))^2dx.\)
(a)
Noting that \(\displaystyle\int (\sin(x))^2dx=\displaystyle\int (\sin(x))(\sin(x))dx\) and letting \(u=\sin(x), dv=\sin(x)dx\text{,}\) what equality does integration by parts yield?
- \(\displaystyle \displaystyle\int (\sin(x))^2dx=\sin(x)\cos(x)+\int (\cos(x))^2 dx.\)
- \(\displaystyle \displaystyle\int (\sin(x))^2dx=-\sin(x)\cos(x)+\int (\cos(x))^2 dx.\)
- \(\displaystyle \displaystyle\int (\sin(x))^2dx=\sin(x)\cos(x)-\int (\cos(x))^2 dx.\)
- \(\displaystyle \displaystyle\int (\sin(x))^2dx=-\sin(x)\cos(x)-\int (\cos(x))^2 dx.\)
(b)
Using the fact that \((\cos(x))^2=1-(\sin(x))^2\) to rewrite the above equality.
(c)
Solve algebraically for \(\displaystyle\int (\sin(x))^2dx.\)
Activity 5.2.21.
Modifying the approach from Activity 5.2.20, use parts to find \(\displaystyle\int (\cos(x))^2dx.\)
