Main content
AP®︎/College Calculus BC
Course: AP®︎/College Calculus BC > Unit 6
Lesson 13: Using integration by parts- Integration by parts intro
- Integration by parts: ∫x⋅cos(x)dx
- Integration by parts: ∫ln(x)dx
- Integration by parts: ∫x²⋅𝑒ˣdx
- Integration by parts: ∫𝑒ˣ⋅cos(x)dx
- Integration by parts
- Integration by parts: definite integrals
- Integration by parts: definite integrals
- Integration by parts challenge
- Integration by parts review
© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice
Integration by parts: ∫x⋅cos(x)dx
This video shows how to find the antiderivative of x*cos(x) using integration by parts. It assigns f(x)=x and g'(x)=cos(x), making f'(x)=1 and g(x)=sin(x). The formula becomes x*sin(x) - ∫sin(x)dx, which simplifies to x*sin(x) + cos(x) + C. Created by Sal Khan.
Want to join the conversation?
Sal, firstly thanks for all these videos. Im really getting alot from them!
Secondly, in the final solution for this video, why is the Constant C not negative? Shouldn't the subtraction reverse the sign? I got: x*sinx+cosx-C.
Correct me if Im wrong please. I'm always making errors on the signs and I want to understand why(65 votes)
The sign for C doesn't really matter as much to the solution of the problem because either way you will get the right equation. Because C is just a constant of integration it is usually put as +C because if the constant is supposed to be negative then you will get C= -3 and if you put -C you will get C=3, which are both the same answers. Not sure if you got it, I explained it as well as I could.(15 votes)
Is integration by parts same thing as using [uv - INTEGRAL(vdu)]?(18 votes)
Yes. The very same if you observe it carefully enough.(39 votes)
At3:30and onwards, how come the 1 in "anti-deriv: 1+sinx" isnt accounted for? shouldn't its anti-derivative be x?(11 votes)- It's not 1+sin(x), it's 1*sin(x). Which equals sin(x). Or 1*1*1*1*1*1*1*1*sin(x), depending on weather.(48 votes)
What is the antiderivative of x ?(0 votes)
or (x^2)/2 - c(2 votes)
Would you still get the same answer even if you assigned the more complicated values to f(x) and the simpler values to g(x)?(6 votes)
I'm not exactly sure what you mean by "values." Are you asking if you'd get the same answer if you switched your choices of function for f(x) and g'(x)? The short answer is no--it basically results in a more complicated integral than the one you started with (Sal references this around1:15-1:30), which makes the problem worse! So you do have to be careful with your choices. The good news is that if you choose the wrong functions (which will be readily apparent when you see the resulting integral), then you can just go back and choose the other combination. Does that help?(13 votes)
I have seen techniques to solve integration , but i am still confused about use of integration . what is use of it !(3 votes)- There are so many uses
There are some great examples here:
http://www.intmath.com/applications-integration/applications-integrals-intro.php
When I am not helping on Khan, I like to work in my recording and music production studio. An effect that gets used a lot in music production is called reverb. Reverb is essentially echos, but the echos are so close together, your ear hears them as one sound and not as distinct replications. The current state of reverb technology now allows you to take a special type of sample, called an impulse response, of an environment you for which you like the reverb, for example, the Sydney Opera House and then, using a process called convolution, a signal, such as a voice that was recorded without reverb, can be transformed by the convolution process using the Sydney Opera House impulse response to produce a voice that has the exact same reverb quality as if it were recorded there! And, you may have guessed it, the convolution algorithm uses integration!
https://en.wikipedia.org/wiki/Convolution#Definition
Here is the website of the makers of the reverb I like to use: https://www.audioease.com/altiverb/
So for me, the process of integration is used on a daily basis!(14 votes)
- What do we do if it is the anti-derivative of xcos5xdx? Would the answer be xsinx + cos5x + C?(2 votes)
- it would be (1/5)xsin5x + (1/25)cos5x + C.
If we assign f(x) to x and g'(x) to cos5x then f(x) is x, f'(x) is 1, g(x) is (1/5)sin5x, and g'(x) is cos5x. g(x) is (1/5)*sin5x because the derivative of that is 5(1/5)cos5x which is just cos5x, the original g'(x). Therefore, when we plug it all back into the formula, we get x(1/5)sin5x - antiderivative of (1(1/5)*sin5x). The antiderivative of (1/5)*sin5x is just (1/25)*-cos5x by the same method that I used earlier. After evaluating the antiderivative, I get (1/5)xsin5x + (1/25)cos5x + C.(10 votes)
- At1:46, isnt the antiderivative of cos(x), sin(x)+c ? Why didn't Sal added the constant ?
Thank-you for your answer(3 votes)- Technically, there is an arbitrary constant at this point, but it is never used in practice. The reason is that there is still an indefinite integral in the expression. When you perform that integration, you will get another arbitrary constant. The sum of two arbitrary constants is again an arbitrary constant, so the one at1:46would be unnecessary.(9 votes)
- Can Integration by Parts be used whenever the integral of the product of any two functions have to be determined ?
Such as can the integral of the expression x√(2x+3) be found by integration by parts ?(3 votes)- Yes, but it is not necessarily the easiest method. Sometimes, no matter what you pick as f and what you pick as g, you won't get something easy to integrate.
For more difficult problems, you may have to apply multiple techniques -- you might need to use parts, u sub and trig sub all mixed together.
So, unfortunately, there is not a simple, works every time, procedure. In fact, there are some seemingly simple-looking integrals that no one has ever solved.
For example, ∫ (x^x) dx has no known solution with a finite number of terms.(5 votes)
Mr. sal how do I integrate some thing complicated like
f(x)=(x²+5x+2)^7 * (3x²+4x-3)^5 ?(2 votes)- Yes, you do have to expand it out. There isn't a nifty trick to get around that, AFAIK. I have done the expansion, but not the integration. Here is what (x²+5x+2)^7 * (3x²+4x-3)^5 expands out to:
243x^24 + 10125x^23 + 190782x^22 + 2146320x^21 + 16021392x^20 + 83306224x^19 + 307462580x^18 + 801041036x^17 + 1415351830x^16 + 1484141906x^15 + 365199968x^14 - 1298344660x^13 - 1744602848x^12 - 491343800x^11 + 752887052x^10 + 682621828x^9 + 25150015x^8 - 208849223x^7 - 79943230x^6 + 17032052x^5 + 17080584x^4 + 2251440x^3 - 959904x^2 - 336960x - 31104(6 votes)
3:30Video transcript
In the last video, I
claimed that this formula would come handy for
solving or for figuring out the antiderivative of
a class of functions. Let's see if that
really is the case. So let's say I want to take
the antiderivative of x times cosine of x dx. Now if you look at this
formula right over here, you want to assign part of this
to f of x and some part of it to g prime of x. And the question is, well do I
assign f of x to x and g prime of x to cosine of x or
the other way around? Do I make f of x cosine
of x and g prime of x, x? And that thing to
realize is to look at the other part of
the formula and realize that you're essentially going
to have to solve this right over here. And here where we
have the derivative of f of x times g of x. So what you want to
do is assign f of x so that the derivative
of f of x is actually simpler than f of x. And assign g prime
of x that, if you were to take its antiderivative,
it doesn't really become any more complicated. So in this case,
if we assign f of x to be equal to x, f prime
of x is definitely simpler, f prime of x is equal to 1. If we assign g prime of x to be
cosine of x, once again, if we take its antiderivative,
that sine of x, it's not any more complicated. If we did it the
other way around, if we set f of x to
be cosine of x, then we're taking its
derivative here. That's not that much
more complicated. But if we set g prime
of x equaling to x and then we had to take
its antiderivative, we get x squared over 2,
that is more complicated. So let me make it
clear over here. We are assigning f of
x to be equal to x. And that means that
the derivative of f is going to be equal to 1. We are assigning-- I'll
write it right here-- g prime of x to be equal to
cosine of x, which means g of x is equal to sine of x,
the antiderivative of cosine of x. Now let's see, given
these assumptions, let's see if we can
apply this formula. So this has all of this. Let's see, the right-hand
side says f of x times g of x. So f of x is x. g of x is sine of x. And then from that, we are going
to subtract the antiderivative of f prime of x-- well,
that's just 1-- times g of x, times sine of x dx. Now this was a huge
simplification. Now I went from trying to solve
the antiderivative of x cosine of x to now I just have
to find the antiderivative of sine of x. And we know the
antiderivative of sine of x dx is just equal to
negative cosine of x. And of course, we can
throw the plus c in now, now that we're pretty
done with taking all of our antiderivatives. So all of this is
going to be equal to x sine of x, x times sine of
x, minus the antiderivative of this, which is just
negative cosine of x. And then we could throw in a
plus c right at the end of it. And doesn't matter if we
subtract a c or add the c. We're saying this is
some arbitrary constant which could even be negative. And so this is all going
to be equal to-- we get our drum roll now-- it's
going to be x times sine of x, subtract a negative, that
becomes a positive, plus cosine of x plus c. And we are done. We were able to take the
antiderivative of something that we didn't know how to take
the antiderivative of before. That was pretty interesting.