Be careful that you know the difference between integration by substitution and integration by parts.
Integration by parts is an extremely important and useful integration technique. Most integrals you come across won't be in a simple form. The idea is to 'reduce' or alter the original integral by breaking it up into pieces that can then be evaluated using the techniques you know so far. You will find that even a small change in an integral will often make a big change in the technique used and the final solution.
If you want a full lecture on this topic, we recommend this video.
video by Prof Leonard |
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WARNING! There is a technique out there that you will see in videos and that some teachers teach (GASP!) called the Table Method. Unless your instructor clearly states that you may use that method, do not UNDER ANY CIRCUMSTANCES use that method. It is a short-cut that will handicap you and keep you from actually learning integration by parts. In fact, we strongly recommend that you learn the technique like we show on this page, even if your instructor teaches you the table method. Then you will be ready in calculus 3 to use the correct technique when you are asked to.
However, as usual, check with your instructor to see what they require.
Where Integration by Parts Comes From
Integration by parts refers to the use of the equation \(\int{ u~dv } = uv - \int{ v~du }\).
It is important to read the next section to understand where this comes from.
The integration by parts equation comes from the product rule for derivatives. We will show an informal proof here.
Let u and v be functions of t.
Using the product and chain rules, we take the derivative of \(u \cdot v\). |
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\(\displaystyle{ \frac{d}{dt}[ u \cdot v] = u \frac{dv}{dt} + v \frac{du}{dt} }\) |
Now we isolate the \(\displaystyle{ u \frac{dv}{dt} }\) term. |
\(\displaystyle{ u \frac{dv}{dt} = \frac{d}{dt}[ u \cdot v] - v \frac{du}{dt} }\) |
Next, we integrate both sides with respect to \(t\) |
\(\displaystyle{ \int{ u \frac{dv}{dt} ~dt } = }\) \(\displaystyle{ \int{ \frac{d}{dt}[ u \cdot v] ~dt } - \int{ v \frac{du}{dt} ~dt } }\) |
Using the idea of differentials, we end up with |
\(\displaystyle{ \int{ u ~dv } = u \cdot v - \int{ v ~du } }\) |
Here is a video showing the same derivation as above.
video by PatrickJMT |
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Using Integration By Parts
The key to using integration by parts is learning what to choose for \(u\) and \(dv\) from the given integral. This is something you will develop a feel for as you get experience working these problems. However, there are some guidelines you can use as you are learning.
General Guidelines
1. Choose the \(dv\) term to make integration easy to get \(v\) since integration is usually more difficult than taking the derivative.
2. If you have a natural log term, always choose it to be equal to u. Remember, \((\ln x)'=1/x\) but \(\int{\ln(x)~dx} \neq 1/x \). [ one of the practice problems solves the integral \(\int{\ln(x)~dx}\) demonstrating what happens when you don't do this ]
3. If you have a polynomial, set it equal to \(u\) since taking the derivative to get du will reduce the power by one. (The reduction formula video below shows why this works.)
LIATE Rule | |
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L | logarithmic functions like \(\ln(x)\) |
I | inverse trig functions like \(\arcsin(x)\) |
A | algebraic functions like polynomials |
T | trigonometric functions like \(\sin(x)\) |
E | exponential functions like \(e^{3x}\) |
source: wikipedia |
Specific Guidelines
There is a rule of thumb called the LIATE rule that suggests which function to choose for \(u\). By going down the list, try choosing the first function you come across as \(u\). This rule does not work ALL the time but it is a good place to start as you are learning integration by parts.
Other Situations That You Will See
- You will come across an integral where you may need to use integration by parts multiple times. This is especially true when you have a polynomial. (The reduction formula video below demonstrates this.)
- Another case you will run into is that, after using integration by parts once or more, you may end up with the same integral you started out with. It looks like you don't accomplish anything or it loops back on itself. However, there is a way to solve these types of integrals. (One of the practice problems solves the integral \(\int{e^{2x}\sin(x)~dx}\) showing this.)
Reduction Formulas
Here are three videos showing how to derive reduction formulas. Although you may never use the results of these videos directly, they are very good videos to watch to help you understand integration by parts.
In the first video, he derives the reduction formula for \( \int{ x^n e^x ~dx} \) and, in the process, demonstrates two of the main points discussed above, (1) choosing \(u\) as the polynomial reduces the power in the next step, and (2) doing integration by parts multiple times is sometimes necessary.
video by Dr Chris Tisdell |
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In this video, she derives the reduction formula for \( \int{ (\ln x)^n ~dx} \).
video by Krista King Math |
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In this video, he derives the reduction formula for \( \int{ \sec^n(x) ~dx} \).
Although the notation gets a bit messy, this is a good video demonstrating the use of integration by parts. If you get lost, just try to get the main points he is making.
video by PatrickJMT |
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Okay, time for you to try some practice problems.
Practice
Unless otherwise instructed, use integration by parts to evaluate these integrals, giving your answers in exact form.
Basic
\(\displaystyle{ \int{x\sin(x)~dx} }\)
Problem Statement |
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Use integration by parts to evaluate \(\displaystyle{ \int{x\sin(x)~dx} }\) giving your answer in exact, simplified, factored form.
Final Answer |
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\(\displaystyle{ \int{x\sin(x)~dx} = \sin(x)-x\cos(x)+C }\)
Problem Statement |
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Use integration by parts to evaluate \(\displaystyle{ \int{x\sin(x)~dx} }\) giving your answer in exact, simplified, factored form.
Solution |
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The general form of integration by parts is \( \int{u~dv} = uv - \int{v~du} \). The key is to choose \( u \) and \( dv \).
\( u = x \) | \( dv = \sin(x)~dx\) | |
\(du = dx\) | \( v = -\cos(x) \) |
We chose \( u = x \) since, after taking the derivative, the power is reduced by one to zero, which causes the \(x\) to disappear. Now we have
\( \int{x \sin(x)~dx} \) |
\( -x \cos(x) - \int{-\cos(x)~dx} \) |
\( -x \cos(x) + \sin(x) + C \) |
\( \sin(x) - x \cos(x) + C \) |
video by Krista King Math |
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Final Answer |
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\(\displaystyle{ \int{x\sin(x)~dx} = \sin(x)-x\cos(x)+C }\)
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\(\displaystyle{ \int{ x \sin(3x) ~ dx } }\)
Problem Statement |
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\(\displaystyle{ \int{ x \sin(3x) ~ dx } }\)
Solution |
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video by Michael Penn |
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\(\displaystyle{ \int{x~\sec^2(x)~dx} }\)
Problem Statement |
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Use integration by parts to evaluate \(\displaystyle{ \int{x~\sec^2(x)~dx} }\) giving your answer in exact, simplified, factored form.
Final Answer |
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\( \int{x \sec^2(x)~dx} = x\tan(x)+\ln|\cos(x)|+C \)
Problem Statement |
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Use integration by parts to evaluate \(\displaystyle{ \int{x~\sec^2(x)~dx} }\) giving your answer in exact, simplified, factored form.
Solution |
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The general form of integration by parts is \( \int{u~dv} = uv - \int{v~du} \).
\( u = x \) | \(dv=\sec^2(x)dx\) | |
\(du = dx\) | \( v = \tan(x) \) |
\(\displaystyle{\int{x \sec^2(x)~dx} }\) |
\(\displaystyle{ x \tan(x) - \int{\tan(x)~dx} }\) |
\(\displaystyle{ x \tan(x) - \int{\frac{\sin(x)}{\cos(x)}dx}}\) |
Now we use substitution by letting \( u = \cos(x) ~~~ \to ~~~ du = -\sin(x)~dx \) |
\(\displaystyle{ \int{\frac{\sin(x)}{\cos(x)}dx} }\) |
\(\displaystyle{ \int{\frac{-1}{u}du} }\) |
\(\displaystyle{ -\ln|u|+C = -\ln|\cos(x)| + C }\) |
Final Answer |
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\( \int{x \sec^2(x)~dx} = x\tan(x)+\ln|\cos(x)|+C \)
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\(\displaystyle{ \int{xe^{-x}~dx} }\)
Problem Statement |
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Use integration by parts to evaluate \(\displaystyle{ \int{xe^{-x}~dx} }\) giving your answer in exact, simplified, factored form.
Solution |
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video by PatrickJMT |
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\(\displaystyle{ \int{ x~5^x~dx } }\)
Problem Statement |
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Use integration by parts to evaluate \(\displaystyle{ \int{ x~5^x~dx } }\) giving your answer in exact, simplified, factored form.
Solution |
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video by PatrickJMT |
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\(\displaystyle{ \int{ x \ln(x) ~ dx } }\)
Problem Statement |
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\(\displaystyle{ \int{ x \ln(x) ~ dx } }\)
Solution |
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video by Michael Penn |
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\(\displaystyle{ \int{\ln(x)~dx} }\)
Problem Statement |
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Use integration by parts to evaluate \(\displaystyle{ \int{\ln(x)~dx} }\) giving your answer in exact, simplified, factored form.
Hint |
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Think of the integrand as \((1)\ln(x)\) and work it similar to the previous problem.
Problem Statement |
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Use integration by parts to evaluate \(\displaystyle{ \int{\ln(x)~dx} }\) giving your answer in exact, simplified, factored form.
Final Answer |
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\(\displaystyle{ \int{\ln x~dx} = x\ln x-x+C }\)
Problem Statement |
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Use integration by parts to evaluate \(\displaystyle{ \int{\ln(x)~dx} }\) giving your answer in exact, simplified, factored form.
Hint |
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Think of the integrand as \((1)\ln(x)\) and work it similar to the previous problem.
Solution |
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Here are 3 video solutions to this problem from 3 different instructors.
video by The Lazy Engineer |
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video by PatrickJMT |
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video by The Organic Chemistry Tutor |
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Final Answer |
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\(\displaystyle{ \int{\ln x~dx} = x\ln x-x+C }\)
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\(\displaystyle{ \int_{1}^{2}{ \frac{\ln(x)}{x^2}~dx } }\)
Problem Statement |
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Use integration by parts to evaluate \(\displaystyle{ \int_{1}^{2}{ \frac{\ln(x)}{x^2}~dx } }\) giving your answer in exact, simplified, factored form.
Solution |
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The first video solves the indefinite integral. The second one solves the given definite integral.
video by The Organic Chemistry Tutor |
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video by PatrickJMT |
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\(\displaystyle{ \int{ x^3\ln(x)~dx } }\)
Problem Statement |
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Use integration by parts to evaluate \(\displaystyle{ \int{ x^3\ln(x)~dx } }\) giving your answer in exact, simplified, factored form.
Solution |
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video by Krista King Math |
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\(\displaystyle{ \int{ x\cos(3x)~dx } }\)
Problem Statement |
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Use integration by parts to evaluate \(\displaystyle{ \int{ x\cos(3x)~dx } }\) giving your answer in exact, simplified, factored form.
Solution |
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video by Krista King Math |
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\(\displaystyle{ \int{ xe^{2x}dx } }\)
Problem Statement |
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Use integration by parts to evaluate \(\displaystyle{ \int{ xe^{2x}dx } }\) giving your answer in exact, simplified, factored form.
Solution |
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video by Krista King Math |
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\(\displaystyle{ \int{ \arctan(x)~dx } }\)
Problem Statement |
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Use integration by parts to evaluate \(\displaystyle{ \int{ \arctan(x)~dx } }\) giving your answer in exact, simplified, factored form.
Solution |
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video by Krista King Math |
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\(\displaystyle{ \int{ xe^x~dx } }\)
Problem Statement |
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Use integration by parts to evaluate \(\displaystyle{ \int{ xe^x~dx } }\) giving your answer in exact, simplified, factored form.
Solution |
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video by Krista King Math |
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\(\displaystyle{ \int{x^7\ln(x)~dx } }\)
Problem Statement |
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Use integration by parts to evaluate \(\displaystyle{ \int{x^7\ln(x)~dx } }\) giving your answer in exact, simplified, factored form.
Solution |
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video by Krista King Math |
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Intermediate
\(\displaystyle{ \int{x^2\sin(x)~dx } }\)
Problem Statement |
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Use integration by parts to evaluate \(\displaystyle{ \int{x^2\sin(x)~dx } }\) giving your answer in exact, simplified, factored form.
Final Answer |
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\(\displaystyle{ \int{x^2\sin(x)~dx} = (2-x^2)\cos(x)+2x\sin(x)+C }\)
Problem Statement |
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Use integration by parts to evaluate \(\displaystyle{ \int{x^2\sin(x)~dx } }\) giving your answer in exact, simplified, factored form.
Solution |
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This solution will require the use of integration by parts twice (due to the \(x^2\) term). The general form is \( \int{u~dv} = uv - \int{v~du} \). The key is to choose \( u \) and \( dv \).
\( u = x^2 \) | \( dv = \sin(x)~dx \) | |
\(du = 2x~dx\) | \( v = -\cos(x) \) |
We chose \(u=x^2 \) since after taking the derivative, the power is reduced by one. Now we have
\(\displaystyle{ \int{x^2 \sin(x)~dx} }\) |
\(\displaystyle{ -x^2 \cos(x) - \int{-\cos(x) (2x~dx)} }\) |
\(\displaystyle{ -x^2 \cos(x) + 2 \int{ x~\cos(x)~dx } }\) |
Now we need to use integration by parts again.
\( u = x \) | \( dv = \cos(x)~dx \) | |
\(du = dx\) | \( v = \sin(x) \) |
Now we have
\( -x^2 \cos(x) + 2 \int{x~\cos(x)~dx} \) |
\( -x^2 \cos(x) + 2 \left[x \sin(x) - \int{\sin(x)~dx}\right] \) |
\( -x^2 \cos(x) + 2x \sin(x) - 2 (-\cos(x)) + C \) |
\( -x^2 \cos(x) + 2x \sin(x) + 2\cos(x) + C \) |
\( (2-x^2)\cos(x) + 2x \sin(x) + C \) |
Final Answer |
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\(\displaystyle{ \int{x^2\sin(x)~dx} = (2-x^2)\cos(x)+2x\sin(x)+C }\)
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\(\displaystyle{\int{x^2\sin^2(x)~dx}}\)
Problem Statement |
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Use integration by parts to evaluate \(\displaystyle{\int{x^2\sin^2(x)~dx}}\) giving your answer in exact, simplified, factored form.
Final Answer |
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\(\displaystyle{ \int{x^2~\sin^2(x)~dx}= }\) \(\displaystyle{ \frac{1}{24}\left[4x^3+3(1-2x^2)\sin(2x)-6x~\cos(2x)\right]+C }\)
Problem Statement |
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Use integration by parts to evaluate \(\displaystyle{\int{x^2\sin^2(x)~dx}}\) giving your answer in exact, simplified, factored form.
Solution |
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This solution will require the use of integration by parts twice (due to the \(x^2\) term). But first let's reduce the \(\sin^2(x)\) using the identity \( \sin^2(x) = (1-\cos(2x))/2 \). Now we have
\(\displaystyle{ \int{x^2~\sin^2(x)~dx} }\) |
\(\displaystyle{ \int{x^2 \frac{1-\cos(2x)}{2} dx} }\) |
\(\displaystyle{ \frac{1}{2}\int{x^2 - x^2 \cos(2x) dx} }\) |
\(\displaystyle{ \frac{1}{2}\left[ \frac{x^3}{3} - \int{x^2\cos(2x)~dx} \right] }\) |
Okay, so let's do integration by parts.
The general form is \( \int{u~dv} = uv - \int{v~du} \). The key is to choose \( u \) and \( dv \).
\( u = x^2 \) | \( dv = \cos(2x)~dx \) | |
\(du = 2x~dx \) | \( v = \sin(2x)/2\) |
We chose \( u = x^2 \) since after taking the derivative, the power is reduced by one. Now we have
\(\displaystyle{ \frac{1}{2}\left[ \frac{x^3}{3} - \int{x^2\cos(2x)~dx} \right] }\) |
\(\displaystyle{ \frac{1}{2}\left[ \frac{x^3}{3} - \frac{x^2}{2}\sin(2x) + \int{\frac{\sin(2x)}{2}(2x)dx } \right] }\) |
\(\displaystyle{ \frac{1}{2}\left[\frac{x^3}{3} - \frac{x^2}{2}\sin(2x) + \int{x~\sin(2x)~dx} \right] }\) |
Now we need to use integration by parts again.
\( u = x \) | \( dv = \sin(2x)~dx \) | |
\(du = dx\) | \( v = -\cos(2x)/2 \) |
Now we have
\(\displaystyle{ \frac{1}{2}\left[ \frac{x^3}{3} - \frac{x^2}{2}\sin(2x) + \int{x~\sin(2x)~dx} \right] }\) |
\(\displaystyle{ \frac{1}{2}\left[ \frac{x^3}{3} - \frac{x^2}{2}\sin(2x) + \frac{-x}{2}\cos(2x) - \int{\frac{-\cos(2x)}{2} dx} \right] }\) |
\(\displaystyle{ \frac{1}{2}\left[ \frac{x^3}{3} - \frac{x^2}{2}\sin(2x) - \frac{x}{2}\cos(2x) + \frac{1}{2}\int{\cos(2x)~dx} \right] }\) |
\(\displaystyle{ \frac{1}{2}\left[ \frac{x^3}{3} - \frac{x^2}{2}\sin(2x) - \frac{x}{2}\cos(2x) + \frac{1}{2}\frac{\sin(2x)}{2} \right] + C }\) |
\(\displaystyle{ \frac{1}{2}\left[ \frac{x^3}{3} - \frac{x^2}{2}\sin(2x) - \frac{x}{2}\cos(2x) + \frac{\sin(2x)}{4} \right] + C }\) |
\(\displaystyle{ \frac{1}{24} \left[ 4x^3 - 6x^2 \sin(2x) - 6x ~\cos(2x) + 3\sin(2x) \right] + C }\) |
\(\displaystyle{ \frac{1}{24}\left[ 4x^3 + 3(1-2x^2)\sin(2x) - 6x~\cos(2x) \right] + C }\) |
Final Answer |
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\(\displaystyle{ \int{x^2~\sin^2(x)~dx}= }\) \(\displaystyle{ \frac{1}{24}\left[4x^3+3(1-2x^2)\sin(2x)-6x~\cos(2x)\right]+C }\)
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\(\displaystyle{ \int{ e^x \sin x ~dx } }\)
Problem Statement |
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Use integration by parts to evaluate \(\displaystyle{ \int{ e^x \sin x ~dx } }\) giving your answer in exact, simplified, factored form.
Solution |
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Here are two solutions to this problem by two different instructors.
video by Michael Penn |
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video by The Organic Chemistry Tutor |
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\(\displaystyle{ \int{ (\ln x)^2dx } }\)
Problem Statement |
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Use integration by parts to evaluate \(\displaystyle{ \int{ (\ln x)^2dx } }\) giving your answer in exact, simplified, factored form.
Solution |
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This problem is solved by two different instructors in these 2 videos.
video by PatrickJMT |
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video by The Organic Chemistry Tutor |
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\(\displaystyle{ \int{ e^{2x}\sin(x)~dx } }\)
Problem Statement |
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Use integration by parts to evaluate \(\displaystyle{ \int{ e^{2x}\sin(x)~dx } }\) giving your answer in exact, simplified, factored form.
Solution |
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video by PatrickJMT |
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\(\displaystyle{ \int{ x^2e^xdx } }\)
Problem Statement |
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Use integration by parts to evaluate \(\displaystyle{ \int{ x^2e^xdx } }\) giving your answer in exact, simplified, factored form.
Solution |
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This is a great video to learn the relationship between integration by parts and the table method. Keep in mind, though, that some instructors may not allow the use of this technique since it is a way to get around understanding how to do integration by parts. As usual, check with your instructor to see what they require.
video by Krista King Math |
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\(\displaystyle{ \int{ e^{7x}\cos(2x)~dx } }\)
Problem Statement |
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Use integration by parts to evaluate \(\displaystyle{ \int{ e^{7x}\cos(2x)~dx } }\) giving your answer in exact, simplified, factored form.
Solution |
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video by Krista King Math |
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\(\displaystyle{ \int{x^3e^x dx } }\)
Problem Statement |
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Use integration by parts to evaluate \(\displaystyle{ \int{x^3e^x dx } }\) giving your answer in exact, simplified, factored form.
Solution |
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video by Krista King Math |
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\(\displaystyle{ \int{ x^3 e^{2x} ~dx } }\)
Problem Statement |
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\(\displaystyle{ \int{ x^3 e^{2x} ~dx } }\)
Solution |
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Although we do not recommend using the tabular method and some instructors do not allow it, you can still use it to check your answer. So we include this video to help you understand it.
video by Michael Penn |
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\(\displaystyle{ \int{ e^{2\theta}\sin(3\theta)~d\theta } }\)
Problem Statement |
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Use integration by parts to evaluate \(\displaystyle{ \int{ e^{2\theta}\sin(3\theta)~d\theta } }\) giving your answer in exact, simplified, factored form.
Solution |
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video by Krista King Math |
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\(\displaystyle{ \int_{ \sqrt{\pi/2}}^{\sqrt{\pi}}{\theta^3 \cos(\theta^2) ~d\theta } }\)
Problem Statement |
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Use integration by parts to evaluate \(\displaystyle{ \int_{ \sqrt{\pi/2}}^{\sqrt{\pi}}{\theta^3 \cos(\theta^2) ~d\theta } }\) giving your answer in exact, simplified, factored form.
Solution |
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Her notation is incorrect here (and you may lose points if you work it like she did). Be careful not to do what she does. She leaves her limits of integration in terms of \(\theta\) when her equations are in terms of x. She should have either dropped her limits of integration while she was using x's or converted her limits in terms of x. So her first integral entirely in terms of \(x\) should have been
\(\displaystyle{ \frac{1}{2}\int_{\pi/2}^{\pi}{x \cos(x)~dx} }\)
Notice that the limits are as follows
\( x = \theta^2 \) | |
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lower limit | \( \theta = \sqrt{\pi/2} \to x = \pi/2 \) |
upper limit | \( \theta = \sqrt{\pi} \to x = \pi \) |
This may seem like a minor issue but it can lead to major problems down the road and you could lose points in class.
video by Krista King Math |
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\(\displaystyle{ \int{ \ln(3r+8)~dr } }\)
Problem Statement |
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Use integration by parts to evaluate \(\displaystyle{ \int{ \ln(3r+8)~dr } }\) giving your answer in exact, simplified, factored form.
Final Answer |
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\( r \ln(3r+8) - r + (8/3)\ln(3r+8) + C \)
Problem Statement |
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Use integration by parts to evaluate \(\displaystyle{ \int{ \ln(3r+8)~dr } }\) giving your answer in exact, simplified, factored form.
Solution |
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video by MIP4U |
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Final Answer |
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\( r \ln(3r+8) - r + (8/3)\ln(3r+8) + C \)
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\(\displaystyle{ \int{ \frac{xe^{2x}}{(1+2x)^2} ~dx } }\)
Problem Statement |
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Use integration by parts to evaluate \(\displaystyle{ \int{ \frac{xe^{2x}}{(1+2x)^2} ~dx } }\) giving your answer in exact, simplified, factored form.
Final Answer |
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\(\displaystyle{ \frac{e^{2x}}{4(1+2x)} + C }\)
Problem Statement |
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Use integration by parts to evaluate \(\displaystyle{ \int{ \frac{xe^{2x}}{(1+2x)^2} ~dx } }\) giving your answer in exact, simplified, factored form.
Solution |
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video by blackpenredpen |
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Final Answer |
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\(\displaystyle{ \frac{e^{2x}}{4(1+2x)} + C }\)
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\(\displaystyle{ \int{ \sin(\ln x) ~dx } }\)
Problem Statement |
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Use integration by parts to evaluate \(\displaystyle{ \int{ \sin(\ln x) ~dx } }\) giving your answer in exact, simplified, factored form.
Solution |
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video by The Organic Chemistry Tutor |
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\(\displaystyle{ \int{ e^{\sqrt{x}} ~ dx } }\)
Problem Statement |
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\(\displaystyle{ \int{ e^{\sqrt{x}} ~ dx } }\)
Hint |
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Start with the substitution \(u=\sqrt{x}\) to get an integral where you can use integration by parts.
Problem Statement |
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\(\displaystyle{ \int{ e^{\sqrt{x}} ~ dx } }\)
Hint |
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Start with the substitution \(u=\sqrt{x}\) to get an integral where you can use integration by parts.
Solution |
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He uses the tabular method in this solution, which we do not recommend.
video by Michael Penn |
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Advanced
\(\displaystyle{ \int{ (4t^2-8t+5)~e^{t^2-3t} dt } }\)
Problem Statement |
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Use integration by parts to evaluate \(\displaystyle{ \int{ (4t^2-8t+5)~e^{t^2-3t} dt } }\) giving your answer in exact, simplified, factored form.
Hint |
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This problem is tricky!
To start, set \(u\) to the exponent. Then find \(du/dt\). Divide the result into the polynomial.
Problem Statement |
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Use integration by parts to evaluate \(\displaystyle{ \int{ (4t^2-8t+5)~e^{t^2-3t} dt } }\) giving your answer in exact, simplified, factored form.
Final Answer |
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\(\displaystyle{\int{(4t^2-8t+5)e^{t^2-3t}dt}=(2t-1)e^{t^2-3t}+C}\)
Problem Statement |
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Use integration by parts to evaluate \(\displaystyle{ \int{ (4t^2-8t+5)~e^{t^2-3t} dt } }\) giving your answer in exact, simplified, factored form.
Hint |
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This problem is tricky!
To start, set \(u\) to the exponent. Then find \(du/dt\). Divide the result into the polynomial.
Solution |
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Let \(u=t^2-3t ~~~ \to ~~~ du/dt=2t-3\) and divide into \(4t^2-8t+5\) to get \(4t^2-8t+5=(2t-1)(2t-3)+2\).
So our integral looks like
\(\displaystyle{\int{ (4t^2-8t+5)e^{t^2-3t} dt } }\) |
\(\displaystyle{\int{ (2t-1)(2t-3)e^{t^2-3t} dt } + \int{ 2e^{t^2-3t} dt }}\) |
Now we will use integration by parts on the first integral.
\( u = 2t-1 \) | \( dv = (2t-3)e^{t^2-3t} dt \) | |
\(du = 2~dt \) | \( v = e^{t^2-3t} \) |
\(\displaystyle{ \int{ (4t^2-8t+5)e^{t^2-3t} dt } }\) |
\(\displaystyle{ \int{ (2t-1)(2t-3)e^{t^2-3t} dt } + \int{ 2e^{t^2-3t} dt } }\) |
\(\displaystyle{ (2t-1)e^{t^2-3t} - \int{2e^{t^2-3t} dt} + \int{ 2e^{t^2-3t} dt } + C }\) |
\(\displaystyle{ (2t-1)e^{t^2-3t} + C }\) |
Final Answer |
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\(\displaystyle{\int{(4t^2-8t+5)e^{t^2-3t}dt}=(2t-1)e^{t^2-3t}+C}\)
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\(\displaystyle{ \int{ \frac{(1-2x)e^{\arctan x}}{(1+x^2)^2} ~ dx } }\)
Problem Statement |
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Use integration by parts to evaluate \(\displaystyle{ \int{ \frac{(1-2x)e^{\arctan x}}{(1+x^2)^2} ~ dx } }\) giving your answer in exact, simplified, factored form.
Hint |
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Let \( u = e^{\arctan x} \).
Problem Statement |
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Use integration by parts to evaluate \(\displaystyle{ \int{ \frac{(1-2x)e^{\arctan x}}{(1+x^2)^2} ~ dx } }\) giving your answer in exact, simplified, factored form.
Hint |
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Let \( u = e^{\arctan x} \).
Solution |
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video by Integrals ForYou |
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Really UNDERSTAND Calculus
related topics on other pages |
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external links you may find helpful |
The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
Set 1 - basic identities | |||
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\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) |
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) |
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) |
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) |
Set 2 - squared identities | ||
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\( \sin^2t + \cos^2t = 1\) |
\( 1 + \tan^2t = \sec^2t\) |
\( 1 + \cot^2t = \csc^2t\) |
Set 3 - double-angle formulas | |
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\( \sin(2t) = 2\sin(t)\cos(t)\) |
\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\) |
Set 4 - half-angle formulas | |
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\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\) |
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) |
Trig Derivatives
\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\) |
\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\) | |
\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\) |
\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\) | |
\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\) |
\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\) |
Inverse Trig Derivatives
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\) |
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\) | |
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) |
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\) | |
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
Trig Integrals
\(\int{\sin(x)~dx} = -\cos(x)+C\) |
\(\int{\cos(x)~dx} = \sin(x)+C\) | |
\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\) |
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\) | |
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) |
\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\) |
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Practice Instructions
Unless otherwise instructed, use integration by parts to evaluate these integrals, giving your answers in exact form.