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## 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

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# Integration by parts: definite integrals

When finding a definite integral using integration by parts, we should first find the antiderivative (as we do with indefinite integrals), but then we should also evaluate the antiderivative at the boundaries and subtract.

## Want to join the conversation?

- Any general rules to know straight away when to use Integration by parts and when to use u-substitution? Thanks.(2 votes)
- Integration by parts tends to be more useful when you are trying to integrate an expression whose factors are different types of functions (e.g. sin(x)*e^x or x^2*cos(x)).

U-substitution is often better when you have compositions of functions (e.g. cos(x)*e^(sin(x)) or cos(x)/(sin(x)^2+1)).(10 votes)

- Does it still work if we chose to assign x to f(x) or g(x) to make it more complicated?(3 votes)
- Yes, but the point is to use the easier way.(7 votes)

- Would this type of question have to clarify if the answer is in DEGREES or RADIANS?(2 votes)
- You must use radians for integration, so your answer should be in radians unless the question explicitly states otherwise for some reason.(2 votes)

- In the next practice exercise, all of the explanations of the correct answers are explained in a different form than this type. Where can I learn the alternate way?(1 vote)
- So the right part of the IBP formula is the antiderivative, right? Is that why you evaluate that whole expression (not just the integral of f'*g) at the limits of integration?(1 vote)
- I am not 100% sure what you mean by right part. If you mean RHS (Right-Hand-Side) then yes.(1 vote)

- The formula of product rule is incorrect as it should be f(x)g'(x) = ∫ f(x)g(x)dx + ∫ g(x)f'(x)dx . i noticed when i was solving for { \int \frac{lnx}{x^2}\:dx } there was a -sign problem in my result , Althouh you guys are doing great job thanks khan acadmey for helping me in learning for free(1 vote)
- If you plug f(x)=1 into your formula, it simplifies to

g'(x) = ∫g(x)dx, which isn't true in general. The formula in the video is correct.(1 vote)

## Video transcript

- [Instructor] We're
gonna do in this video is try to evaluate the
definite integral from zero to pi of x cosine of x dx. Like always, pause this video and see if you can evaluate it yourself. Well when you immediately look at this, it's not obvious how you
just straight up take the anti-derivative here
and then evaluate that at pi and then subtract from that and evaluate it at zero, so
we're probably going to have to use a slightly more
sophisticated technique. And in general, if you
see a product of functions right over here, and if
one of these functions is fairly straightforward to
take the anti-derivative of without making it more
complicated like cosine of x, and another of the functions
like x if you were to take its derivative, it gets simpler. In this case, it would just become one. It's a pretty good sign
that we should be using integration by parts. So let's just remind ourselves
about integration by parts. So integration by parts,
I'll do it right over here, if I have the integral
and I'll just write this as an indefinite integral
but here we wanna take the indefinite integral
and then evaluate it at pi and evaluate it at zero,
so if I have f of x times g prime of x, dx, this is going to be equal
to, and in other videos we prove this, it really
just comes straight out of the product rule that you learned in differential calculus,
this is gonna be equal to f of x times g of x minus,
you then swap these around, minus f prime of x, g of x, dx. And just to reiterate what I said before, you wanna find an f of x that
when I take its derivative, it simplifies it, so simplify, and you wanna find a g
prime of x that when I take its anti-derivative, so when
I take its anti-derivative, it doesn't get more complicated. So not more complicated. Because if the f of x gets
simplified when I take its derivative, and the
g prime of x does not get more complicated when
I take its anti-derivative, then this expression will
maybe be easier to find the anti-derivative of. So let's do that over here. Between x and cosine of x, which one gets more simple when
I take its derivative? Well the derivative of x is
just one, so I'm gonna make that my f of x, so I could
write that over here, so my f of x, I will
say is x, in which case, f prime of x, f prime of x,
is going to be equal to one, and then what would my g prime of x be? Well my g prime of x,
cosine of x if I take its anti-derivative, it
doesn't get more complicated. The anti-derivative of
cosine of x is sine of x. So let me make that my g prime of x. So g prime of x is equal to cosine of x in which case g of x, the
anti-derivative of cosine of x, well it's just sine of x or
another way to think about it, the derivative of sine
of x is cosine of x. Now you could think about
plus C's and all of that but remember, this is gonna
be a definite integral so all of those arbitrary
constants are going to get canceled out. So now let's think through this. Let's just apply the
integration by parts here. In this particular case,
all of this is going to be equal to, so I'm saying
that is equal to this, I'm gonna skip down here,
it's going to be equal to f of x times g of x. So that is f of x is
x, g of x is sine of x, f of x times g of x, minus the integral of f prime of x, f prime of x is just one, we could write it like
that, one times g of x, g of x is sine of x, so we
could write it like this, but one times sine of x, well
we could just rewrite that as sine of x, it'll make
it a little bit simpler, sine of x, dx, and then
remember, this is a definite integral, so we are
going to want to evaluate this whole thing at pi
and at zero, and then take the difference between the two. But what is the indefinite
integral of sine of x dx? Well, or to say the anti-derivative of it, we know that the derivative
of cosine is negative sine of x, and so in fact what
we want, we could bring this negative sine into the
integral, so we could say plus the integral of negative sine of x, now this clearly the anti-derivative here is cosine of x, so this
thing is going to be cosine of x, and now we
just have to evaluate it at the end points. So let's first evaluate
this whole thing at pi. So this is going to be
equal to pi sine of pi, pi sine of pi, plus cosine of pi, and then from that I'm going
to subtract this whole thing evaluated at zero, so let me
do zero in a different color, at zero, so it's going
to be zero times sine of zero plus cosine of zero, so let's see, sine of pi is just zero
so this is just going to cancel out. Cosine of pi, that is negative one, and then this is zero,
and then cosine of zero, that is one, so you have
negative one minus one, so this all gets us to negative two, and we are done. Using integration by parts,
we were able to evaluate this definite integral.