Integration by Parts
Activity 5.2.1
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Using the product rule, which of these is derivative of \(x^3e^x\) with respect to \(x\text{?}\)
\(\displaystyle 3x^2e^x\)
\(\displaystyle 3x^2e^{x}+x^3e^x\)
\(\displaystyle 3x^2e^{x-1}\)
\(\displaystyle \frac{1}{4}x^4 e^x\)
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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
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Which differentiation rule is easier to implement?
Product Rule
Chain Rule
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Which differentiation strategy do expect to be easier to reverse?
Product Rule
Chain Rule
Activity 5.2.3
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Which of the following equations is equivalent to the formula \(\frac{d}{dx}[uv]=u'v+uv'\text{?}\)
\(\displaystyle uv'=-\frac{d}{dx}(uv)-vu'\)
\(\displaystyle uv'=-\frac{d}{dx}(uv)+vu'\)
\(\displaystyle uv'=\frac{d}{dx}(uv)+vu'\)
\(\displaystyle uv'=\frac{d}{dx}(uv)-vu'\)
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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
Activity 5.2.5
Consider \(\int xe^{x}\,dx\text{.}\) Suppose we decided to let \(u=x\text{.}\)
Compute \(\frac{du}{dx}=\unknown\text{,}\) and rewrite it as \(du=\unknown\,dx\text{.}\)
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What is the best candidate for \(dv\text{?}\)
\(\displaystyle dv=x\,dx\)
\(\displaystyle dv=e^x\)
\(\displaystyle dv=x\)
\(\displaystyle dv=e^x\,dx\)
Given that \(dv=e^x\,dx\text{,}\) find \(v=\unknown\text{.}\)
Show why \(\int xe^{x}\,dx\) may now be rewritten as \(xe^x-\int e^x\,dx\text{.}\)
Solve \(\int e^x\,dx\text{,}\) and then give the most general antiderivative of \(\int xe^{x}\,dx\text{.}\)
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 \(dv=x^9\,dx\text{.}\)
What is \(du\text{?}\)
What is \(v\text{?}\)
What do you get when plugging these pieces into integration by parts?
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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 [REF=TODO] ?
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:
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 \(\int t^5 \sin(t^3)\,dt\text{.}\)
Use the substitution \(x=t^3\) to rewrite the integral in terms of \(x\text{.}\)
Use integration by parts to evaluate the integral in terms of \(x\text{.}\)
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 \(\displaystyle\int (\sin(x))^2dx.\)
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?
- \(\displaystyle \displaystyle\int (\sin(x))^2dx=\sin(x)\cos(x)+\int (\cos(x))^2 dx.\)
- \(\displaystyle \displaystyle\int (\sin(x))^2dx=-\sin(x)\cos(x)+\int (\cos(x))^2 dx.\)
- \(\displaystyle \displaystyle\int (\sin(x))^2dx=\sin(x)\cos(x)-\int (\cos(x))^2 dx.\)
- \(\displaystyle \displaystyle\int (\sin(x))^2dx=-\sin(x)\cos(x)-\int (\cos(x))^2 dx.\)
Using the fact that \((\cos(x))^2=1-(\sin(x))^2\) to rewrite the above equality.
Solve algebraically for \(\displaystyle\int (\sin(x))^2dx.\)