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Calculus for Team-Based Inquiry Learning 2025 Edition

TBIL Community

Section 5.2 Integration by Parts (TI2)

Subsection 5.2.1 Activities

Activity 5.2.1.

Answer the following.
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(a)
Using the product rule, which of these is derivative of \(x^3e^x\) with respect to \(x\text{?}\)
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  1. \(\displaystyle 3x^2e^x\)
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  2. \(\displaystyle 3x^2e^{x}+x^3e^x\)
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  3. \(\displaystyle 3x^2e^{x-1}\)
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  4. \(\displaystyle \dfrac{1}{4}x^4 e^x\)
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(b)
Based on this result, which of these would you suspect to equal \(\displaystyle \int 3x^2e^x+x^3e^x\,dx\text{?}\)
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  1. \(\displaystyle x^3e^x+C\)
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  2. \(\displaystyle x^3e^x+\dfrac{1}{4}x^4e^x+C\)
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  3. \(\displaystyle 6xe^x+3x^2e^x+C\)
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  4. \(\displaystyle 6xe^x+3x^2e^x+3x^2e^x+x^3e^x+C\)
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Activity 5.2.3.

Answer the following.
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(a)
Which of the following equations is equivalent to the formula \(\dfrac{d}{dx}[uv]=u'v+uv'\text{?}\)
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  1. \(\displaystyle uv'=-\dfrac{d}{dx}(uv)-vu'\)
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  2. \(\displaystyle uv'=-\dfrac{d}{dx}(uv)+vu'\)
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  3. \(\displaystyle uv'=\dfrac{d}{dx}(uv)+vu'\)
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  4. \(\displaystyle uv'=\dfrac{d}{dx}(uv)-vu'\)
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(b)
Which of these is the most concise result of integrating both sides with respect to \(x\text{?}\)
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  1. \(\displaystyle \displaystyle \int(uv')\,dx=uv-\int(vu')\,dx\)
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  2. \(\displaystyle \displaystyle \int(u)\,dv=uv-\int(v)\,du\)
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  3. \(\displaystyle \displaystyle \int(uv')\,dx=uv-\int(vu')\,dx+C\)
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  4. \(\displaystyle \displaystyle \int(u)\,dv=uv-\int(v)\,du+C\)
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Activity 5.2.5.

Consider \(\displaystyle \int xe^{x}\,dx\text{.}\) Suppose we decided to let \(u=x\text{.}\)
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(a)
Compute \(\dfrac{du}{dx}=\unknown\text{,}\) and rewrite it as \(du=\unknown\,dx\text{.}\)
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(c)
Given that \(dv=e^x\,dx\text{,}\) find \(v=\unknown\text{.}\)
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(d)
Show why \(\displaystyle \int xe^{x}\,dx\) may now be rewritten as \(\displaystyle xe^x-\int e^x\,dx\text{.}\)
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(e)
Solve \(\displaystyle \int e^x\,dx\text{,}\) and then give the most general antiderivative of \(\displaystyle \int xe^{x}\,dx\text{.}\)
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Example 5.2.6.

Here is how one might write out the explanation of how to find \(\displaystyle \int xe^{x}\,dx\) from start to finish:
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\begin{align*} \displaystyle \int xe^{x}\,dx\\ & u=x & dv = e^x\,dx\\ & du=1\cdot\,dx & v=e^x\\ \displaystyle \int xe^x \,dx &= xe^x-\int e^x \,dx\\ &= xe^{x}-e^x+C \end{align*}
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Activity 5.2.7.

Which step of the previous example do you think was the most important?
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  1. Choosing \(u=x\) and \(dv=e^x\,dx\text{.}\)
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  2. Finding \(du=1\,dx\) and \(v=e^x\,dx\text{.}\)
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  3. Applying integration by parts to rewrite \(\displaystyle \int xe^x\,dx\) as \(\displaystyle xe^x-\int e^x\,dx\text{.}\)
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  4. Integrating \(\displaystyle \int e^x\,dx\) to get \(xe^{x}-e^x+C\text{.}\)
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Activity 5.2.8.

Consider the integral \(\displaystyle \int x^9\ln(x) \,dx\text{.}\) Suppose we proceed using integration by parts. We choose \(u=\ln(x)\) and \(dv=x^9\,dx\text{.}\)
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(c)
What do you get when plugging these pieces into integration by parts?
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(d)
Does the new integral \(\displaystyle \int v\,du\) seem easier or harder to compute than the original integral \(\displaystyle \int x^9\ln(x) \,dx\text{?}\)
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  1. The original integral is easier to compute.
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  2. The new integral is easier to compute.
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  3. Neither integral seems harder than the other one.
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Activity 5.2.9.

Consider the integral \(\displaystyle \int x^9\ln(x) \,dx\) 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?
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  1. We prefer the substitution choice of \(u=\ln(x)\) and \(dv=x^9\,dx\text{.}\)
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  2. We prefer the substitution choice of \(u=x^9\) and \(dv=\ln(x)\,dx\text{.}\)
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  3. We do not have a strong preference, since these choices are of the same difficulty.
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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?
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  1. \(u=\cos(x)\text{,}\) \(dv=x\, dx\)
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  2. \(u=\cos(x)\,dx\text{,}\) \(dv=x\)
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  3. \(u=x\,dx\text{,}\) \(dv=\cos(x)\)
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  4. \(u=x\text{,}\) \(dv=\cos(x)\,dx\)
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Activity 5.2.11.

Evaluate the integral \(\displaystyle \int x\cos(x)\,dx\) using integration by parts.
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Activity 5.2.12.

Now use integration by parts to evaluate the integral \(\displaystyle \int_{\pi/6}^{\pi} x\cos(x)\,dx\text{.}\)
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Activity 5.2.13.

Consider the integral \(\displaystyle \int x\arctan(x)\,dx\text{.}\) Suppose we proceed using integration by parts. Which of the following candidates for \(u\) and \(dv\) would best allow you to evaluate this integral?
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  1. \(u=x\,dx\text{,}\) \(dv=\arctan(x)\)
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  2. \(u=\arctan(x)\text{,}\) \(dv=x\,dx\)
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  3. \(u=x\arctan(x)\text{,}\) \(dv=\,dx\)
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  4. \(u=x\text{,}\) \(dv=\arctan(x)\,dx\)
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Activity 5.2.14.

Consider the integral \(\displaystyle \int e^x\cos(x)\,dx\text{.}\) Suppose we proceed using integration by parts. Which of the following candidates for \(u\) and \(dv\) would best allow you to evaluate this integral?
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  1. \(u=e^x\text{,}\) \(dv=\cos(x)\,dx\)
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  2. \(u=\cos(x)\text{,}\) \(dv=e^x\,dx\)
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  3. \(u=e^x\,dx\text{,}\) \(dv=\cos(x)\)
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  4. \(u=\cos(x)\,dx\text{,}\) \(dv=e^x\)
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Activity 5.2.15.

Suppose we started using integration by parts to solve the integral \(\displaystyle \int e^x\cos(x)\,dx\) as follows:
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\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 \(\displaystyle \int e^x\sin(x) \,dx\text{.}\) Which of the following candidates for \(u\) and \(dv\) would best allow you to continue evaluating the original integral \(\displaystyle \int e^x\cos(x)\,dx\text{?}\)
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  1. \(u=e^x\text{,}\) \(dv=\sin(x)\,dx\)
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  2. \(u=\sin(x)\text{,}\) \(dv=e^x\,dx\)
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  3. \(u=e^x\,dx\text{,}\) \(dv=\sin(x)\)
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  4. \(u=\sin(x)\,dx\text{,}\) \(dv=e^x\)
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Activity 5.2.16.

Use integration by parts to show that \(\displaystyle \int_0^{\pi/4} x\sin(2x)\,dx=\dfrac{1}{4}\text{.}\)
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Activity 5.2.17.

Consider the integral \(\displaystyle \int t^5 \sin(t^3)\,dt\text{.}\)
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(a)
Use the substitution \(x=t^3\) to rewrite the integral in terms of \(x\text{.}\)
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(b)
Use integration by parts to evaluate the integral in terms of \(x\text{.}\)
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(c)
Replace \(x\) with \(t^3\) to finish evaluating the original integral.
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Activity 5.2.18.

Use integration by parts to show that \(\displaystyle \int \ln(z)\,dz=z \ln(z) - z + C\text{.}\)
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Activity 5.2.19.

Given that that \(\displaystyle \int \ln(z)\,dz=z \ln(z) - z + C\text{,}\) evaluate \(\displaystyle \int (\ln(z))^2\,dz\text{.}\)
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Activity 5.2.20.

Consider the antiderivative \(\displaystyle\int (\sin(x))^2\, dx.\)
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(a)
Noting that \(\displaystyle\int (\sin(x))^2\, dx=\int (\sin(x))(\sin(x))dx\) and letting \(u=\sin(x), dv=\sin(x)\, dx\text{,}\) what equality does integration by parts yield?
  1. \(\displaystyle \displaystyle\int (\sin(x))^2dx=\sin(x)\cos(x)+\int (\cos(x))^2\, dx.\)
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  2. \(\displaystyle \displaystyle\int (\sin(x))^2dx=-\sin(x)\cos(x)+\int (\cos(x))^2\, dx.\)
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  3. \(\displaystyle \displaystyle\int (\sin(x))^2dx=\sin(x)\cos(x)-\int (\cos(x))^2\, dx.\)
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  4. \(\displaystyle \displaystyle\int (\sin(x))^2dx=-\sin(x)\cos(x)-\int (\cos(x))^2\, dx.\)
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(b)
Use the fact that \((\cos(x))^2=1-(\sin(x))^2\) to rewrite the above equality.
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(c)
Solve algebraically for \(\displaystyle\int (\sin(x))^2\, dx.\)
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Activity 5.2.21.

Modifying the approach from Activity 5.2.20, use parts to find \(\displaystyle\int (\cos(x))^2\, dx.\)
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Subsection 5.2.2 Videos

Figure 105. Video: Compute integrals using integration by parts
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Subsection 5.2.3 Exercises

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