INTEGRATION BY PARTS
Recall: Product Rule
The product rule can be expressed as follows:
𝑑
[𝑓(𝑥)𝑔(𝑥)] = 𝑓(𝑥)𝑔′(𝑥) + 𝑓 ′ (𝑥)𝑔(𝑥)
𝑑𝑥
This implies that 𝑓(𝑥)𝑔(𝑥) is an anti derivative of the function on the right side, hence we can express
this equation in the integral form as follows:
𝑓(𝑥)𝑔(𝑥) = ∫[ 𝑓(𝑥)𝑔′(𝑥) + 𝑓 ′ (𝑥)𝑔(𝑥) ] 𝑑𝑥
𝑓(𝑥)𝑔(𝑥) = ∫ 𝑓(𝑥)𝑔′(𝑥) 𝑑𝑥 + ∫ 𝑓 ′ (𝑥)𝑔(𝑥) 𝑑𝑥
Therefore,
∫ 𝑓(𝑥)𝑔′(𝑥) 𝑑𝑥 = 𝑓(𝑥)𝑔(𝑥) − ∫ 𝑓 ′ (𝑥)𝑔(𝑥) 𝑑𝑥
This formula allows us to compute a difficult integral by computing a much simpler integral. We often
express the integration by part using the following formula:
Let
𝑢 = 𝑓(𝑥) therefore 𝑑𝑢 = 𝑓 ′ (𝑥) 𝑑𝑥
𝑣 = 𝑔(𝑥) therefore 𝑑𝑣 = 𝑔′ (𝑥)𝑑𝑥
Then the formula for integration by parts becomes:
∫ 𝑢 𝑑𝑣 = 𝑢𝑣 − ∫ 𝑣 𝑑𝑢
To integrate by parts we need to strategically choose the u and the dv then apply the formula.
Choose u and dv so that :
 u is very easy to differentiate.
 dv is very easy to integrate.
 ∫ 𝑣 𝑑𝑢 is easier to solve than ∫ 𝑢 𝑑𝑣
Another useful strategy for choosing u and dv is the LIATE method developed by Kasube (1983). This
method can be applied if the integrand is a product of two functions from different categories as below:
Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential
The u will be the function whose category occurs earlier in the list and the second will be the dv. Although
this method does not works every time but often very useful.
1
EXAMPLE 1
∫ 𝑥 sin 𝑥 𝑑𝑥
Solution:
Let us divide the integral into two parts (u and dv), whereby:
𝑢=𝑥
then when differentiate we have
𝑑𝑣 = sin 𝑥 𝑑𝑥 then when integrate we have
𝑑𝑢 = 𝑑𝑥
𝑣 = ∫ sin 𝑥 𝑑𝑥 = cos 𝑥
By applying the formula,
∫ 𝑢 𝑑𝑣 = 𝑢𝑣 − ∫ 𝑣 𝑑𝑢
∫ 𝑥 𝑠𝑖𝑛𝑥 𝑑𝑥 = 𝑥 cos 𝑥 − ∫ sin 𝑥 𝑑𝑥
Here, the integral becomes less complicated to integrate. Then the final answer :
∫ 𝑥 𝑠𝑖𝑛𝑥 𝑑𝑥 = 𝑥 cos 𝑥 − cos 𝑥 + 𝐶
EXAMPLE 2
∫ 𝑥 𝑒 𝑥 𝑑𝑥
Solution:
Let,
𝑢=𝑥
𝑑𝑢 = 𝑑𝑥
𝑑𝑣 = 𝑒 𝑥 𝑑𝑥
𝑣 = ∫ 𝑒 𝑥 𝑑𝑥
𝑣 = 𝑒𝑥
By using the formula,
∫ 𝑥 𝑒 𝑥 𝑑𝑥 = 𝑥𝑒 𝑥 − ∫ 𝑒 𝑥 𝑑𝑥 = 𝑥𝑒 𝑥 − 𝑒 𝑥 + 𝐶
2
Integration involving Natural logarithmic
Recall that :
𝑑
𝑑𝑥
(ln 𝑥 ) =
1
1
. Hence, ∫ 𝑥 𝑑𝑥 = ln 𝑥 + 𝐶 .
𝑥
Therefore there are no special means of
integrating ln x straight away. The only way to integrate function of ln x is by using the integration by
parts.
EXAMPLE 3
∫ 𝑙𝑛 𝑥 𝑑𝑥
Solution:
Let
𝑢 = ln 𝑥
𝑑𝑣 = 𝑑𝑥
1
𝑑𝑢 = 𝑥 𝑑𝑥
𝑣=𝑥
By applying the formula,
∫ 𝑢 𝑑𝑣 = 𝑢𝑣 − ∫ 𝑣 𝑑𝑢
∫ 𝑙𝑛 𝑥 𝑑𝑥 = 𝑥 𝑙𝑛 𝑥 − ∫ 𝑥
1
𝑑𝑥
𝑥
= 𝑥 ln 𝑥 − ∫ 𝑑𝑥
= 𝑥 ln 𝑥 − 𝑥 + 𝐶
Integration involving Inverse Trigonometric
The same case refers when integrating inverse trigonometric function. We cannot directly integrate
inverse trigonometric functions. The techniques of integrating by parts have to be used when
integrating functions involving inverse trigonometry.
EXAMPLE 4
∫ tan−1 𝑥 𝑑𝑥
3
Solution:
Since we cannot directly integrate tan−1 𝑥 , but we can easily differentiate tan−1 𝑥
Therefore,
𝑢 = tan−1 𝑥
𝑑𝑢 =
𝑑𝑣 = 𝑑𝑥
1
𝑑𝑥
1+𝑥 2
𝑣=𝑥
By applying the formula,
∫ 𝑢 𝑑𝑣 = 𝑢𝑣 − ∫ 𝑣 𝑑𝑢
1
∫ tan−1 𝑥 𝑑𝑥 = 𝑥 tan−1 𝑥 − ∫(𝑥) (
) 𝑑𝑥
1 + 𝑥2
= 𝑥 tan−1 𝑥 − ∫
𝑥
𝑑𝑥
1 + 𝑥2
By using basic substitution method,
Let 𝑢 = 1 + 𝑥 2 , then 𝑑𝑢 = 2𝑥 𝑑𝑥 or can be written as
∫
𝑑𝑢
2
= 𝑥𝑑𝑥 . By substituting into the integral,
𝑥
1 𝑑𝑢
1 1
1
1
𝑑𝑥
=
∫
=
∫
𝑑𝑢
=
ln
𝑢
=
ln|1 + 𝑥 2 |
1 + 𝑥2
𝑢 2
2 𝑢
2
2
Hence,
∫ tan−1 𝑥 𝑑𝑥 = 𝑥 tan−1 𝑥 − ∫
= 𝑥 tan−1 𝑥 −
𝑥
𝑑𝑥
1 + 𝑥2
1
ln|1 + 𝑥 2 | + 𝐶
2
𝐨𝐫 𝑥 tan−1 𝑥 − ln √1 + 𝑥 2 + 𝐶
4
Sometimes we also need to do some substitutions to make the integral easy before integration by parts
can be applied.
EXAMPLE 5
∫ 𝑥 cos −1 𝑥 2 𝑑𝑥
Solution:
Let 𝑤 = 𝑥 2 , then 𝑑𝑤 = 2𝑥 𝑑𝑥, therefore 𝑥 𝑑𝑥 =
𝑑𝑤
2
. Hence, the integral becomes,
𝑑𝑤
1
= ∫ cos −1 𝑤 𝑑𝑤
2
2
∫ cos −1 𝑤
Let,
𝑢 = cos−1 𝑤
𝑑𝑢 = −
1
√1−𝑤 2
𝑑𝑣 = 𝑑𝑤
𝑑𝑤
𝑣=𝑤
By applying the integration by parts,
1
1
1
1
∫ cos −1 𝑤 𝑑𝑤 = 𝑤 cos −1 𝑤 − ∫ 𝑤 . −
𝑑𝑤
2
2
2
√1 − 𝑤 2
1
1
𝑤
= 𝑤 cos −1 𝑤 + ∫
𝑑𝑤
2
2 √1 − 𝑤 2
For,
∫
𝑤
√1 − 𝑤 2
𝑑𝑤 = −√1 − 𝑤 2 + 𝐶
(Verify this, by using substitution, 𝑝 = 1 − 𝑤 2)
Hence,
1
1
1
∫ cos−1 𝑤 𝑑𝑤 = 𝑤 cos −1 𝑤 − √1 − 𝑤 2 + 𝐶
2
2
2
Finally, substitute back in terms of x,
∫ 𝑥 cos−1 𝑥 2 𝑑𝑥 =
𝑥2
1
cos−1 𝑥 2 − √1 − 𝑥 2 + 𝐶
2
2
5
EXAMPLE 6
3
∫ 𝑥 5 𝑒 𝑥 𝑑𝑥
Solution:
We cannot directly choose the u and the dv, and then applied the integration by parts. The integral will
become too complicated. (See note below) The easy way here is to use method of substitution, before
integration by parts can be applied. Obviously, we cannot choose u or v when substituting, because of
the formula by parts itself already have the variable u and v. You can choose any variable other than u
and v so that the integral will be less confusing!
Let 𝑝 = 𝑥 3 , then 𝑑𝑝 = 3𝑥 2 𝑑𝑥. Then
𝑑𝑝
3
= 𝑥 2 𝑑𝑥. By splitting the powers of x we have,
3
∫ 𝑥 3 𝑥 2 𝑒 𝑥 𝑑𝑥
By substituting, we have the following integral in terms of p,
1
∫ 𝑝 𝑒 𝑝 𝑑𝑝
3
From here, it is much simpler to use integration by parts.
Let,
𝑑𝑣 = 𝑒 𝑝 𝑑𝑝
𝑢=𝑝
𝑣 = 𝑒𝑝
𝑑𝑢 = 𝑑𝑝
Therefore,
1
1
∫ 𝑝 𝑒 𝑝 𝑑𝑝 = (𝑝 𝑒 𝑝 − ∫ 𝑒 𝑝 𝑑𝑝)
3
3
1
1
= 𝑝 𝑒𝑝 − 𝑒𝑝 + 𝐶
3
3
Hence, by substituting back in terms of x,
3
∫ 𝑥 5 𝑒 𝑥 𝑑𝑥 =
1 3 𝑥3 1 𝑥 3
𝑥 𝑒 − 𝑒 +𝐶
3
3
3
Note : This example shows how to integrate when there is a function involving 𝑒 𝑥 or the similar forms. You have
to figure out critically on how to go about the function to make it less complicated to integrate! Normally basic
2
substitution can be applied here. Note that we cannot directly integrate the function of 𝑒 𝑥 . This function is famous
example of a function that has no anti derivative i.e. we cannot integrate the function!
6
Special Cases of Integration by Parts
There is a special case when integrating functions in terms of the product of exponent and sine/cosine,
for example integrating : 𝑒 𝑥 cos 𝑥 or 𝑒 𝑥 sin 𝑥.
EXAMPLE 7
∫ 𝑒 𝑥 cos 𝑥 𝑑𝑥
Solution:
Let,
𝑑𝑣 = 𝑒 𝑥 𝑑𝑥
𝑢 = cos 𝑥
𝑣 = 𝑒𝑥
𝑑𝑢 = − sin 𝑥 𝑑𝑥
∫ 𝑒 𝑥 cos 𝑥 𝑑𝑥 = 𝑒 𝑥 cos 𝑥 − ∫(𝑒 𝑥 )(− sin 𝑥 )𝑑𝑥
= 𝑒 𝑥 cos 𝑥 + ∫ 𝑒 𝑥 sin 𝑥 𝑑𝑥
Here, we have to again apply the integration by parts,
Let,
𝑢 = sin 𝑥
𝑑𝑣 = 𝑒 𝑥 𝑑𝑥
𝑑𝑢 = cos 𝑥 𝑑𝑥
𝑣 = 𝑒𝑥
Therefore the integral becomes,
∫ 𝑒 𝑥 cos 𝑥 = 𝑒 𝑥 cos 𝑥 + 𝑒 𝑥 sin 𝑥 − ∫ 𝑒 𝑥 cos 𝑥 𝑑𝑥
Here, we can see again the∫ 𝑒 𝑥 cos 𝑥 𝑑𝑥 . Let us bring ∫ 𝑒 𝑥 cos 𝑥 𝑑𝑥 to the left side of the equation,
hence the integral will become :
2 ∫ 𝑒 𝑥 cos 𝑥 𝑑𝑥 = 𝑒 𝑥 cos 𝑥 + 𝑒 𝑥 sin 𝑥
Finally, we divide both sides of the equation by 2, then the final answer :
1
1
∫ 𝑒 𝑥 cos 𝑥 𝑑𝑥 = 𝑒 𝑥 cos 𝑥 + 𝑒 𝑥 sin 𝑥 + 𝐶
2
2
Note : You don’t have to divide the constant C by 2, this is because C is just an arbitrary constant of the integration.
Normally we put C at the final answer of the integration by parts.
7
EXAMPLE 8
By using integration by parts, show that :
∫ 𝑐𝑜𝑠 𝑛 𝑥 𝑑𝑥 =
1
𝑛−1
𝑐𝑜𝑠 𝑛−1 𝑥 𝑠𝑖𝑛 𝑥 +
∫ 𝑐𝑜𝑠 𝑛−2 𝑥 𝑑𝑥
𝑛
𝑛
Solution:
Verify that :
∫ cosn 𝑥 𝑑𝑥 = ∫ cos n−1 𝑥 cos 𝑥 𝑑𝑥
Then let,
𝑢 = cos n−1 𝑥
𝑑𝑣 = cos 𝑥 𝑑𝑥
𝑑𝑢 = (𝑛 − 1)(cos 𝑥)𝑛−2 (− sin 𝑥) 𝑑𝑥
𝑣 = sin 𝑥
= −(𝑛 − 1) cos 𝑛−2 𝑥 sin 𝑥 𝑑𝑥
Hence,
∫ cos n 𝑥 𝑑𝑥 = ∫ cosn−1 𝑥 cos 𝑥 𝑑𝑥 = cos𝑛−1 𝑥 sin 𝑥 + (𝑛 − 1) ∫ sin2 𝑥 cos 𝑛−2 𝑥 𝑑𝑥
= cos𝑛−1 𝑥 sin 𝑥 + (𝑛 − 1) ∫(1 − cos 2 𝑥) cos 𝑛−2 𝑥 𝑑𝑥
= cos 𝑛−1 𝑥 sin 𝑥 + (𝑛 − 1) ∫ cos 𝑛−2 𝑥 𝑑𝑥 + (𝑛 − 1) ∫ cos 𝑛 𝑥 𝑑𝑥
Here we see again the integral of cos𝑛 𝑥 on the RHS. We bring this integrand on the LHS, we get,
𝑛 ∫ cos𝑛 𝑥 𝑑𝑥 = cos𝑛−1 𝑥 sin 𝑥 + (𝑛 − 1) ∫ cos𝑛−2 𝑥 𝑑𝑥
Then by dividing on both sides by n, it is shown that,
∫ 𝑐𝑜𝑠 𝑛 𝑥 𝑑𝑥 =
1
𝑛−1
𝑐𝑜𝑠 𝑛−1 𝑥 𝑠𝑖𝑛 𝑥 +
∫ 𝑐𝑜𝑠 𝑛−2 𝑥 𝑑𝑥
𝑛
𝑛
8
Definite Integration by Parts
A definite integral can be evaluated only after the appropriate integration has been performed. For
evaluating definite integral, the term uv must be evaluated between [a, b].
𝑏
𝑏
∫ 𝑢 𝑑𝑣 = [𝑢𝑣]𝑏𝑎 − ∫ 𝑣 𝑑𝑢
𝑎
𝑎
EXAMPLE 9
1
∫ 𝑥 𝑒 𝑥 𝑑𝑥
0
Solution:
Let,
𝑑𝑣 = 𝑒 𝑥 𝑑𝑥
𝑢=𝑥
𝑣 = ∫ 𝑒 𝑥 𝑑𝑥
𝑑𝑢 = 𝑑𝑥
𝑣 = 𝑒𝑥
1
1
∫ 𝑥 𝑒 𝑥 𝑑𝑥 = [𝑥 𝑒 𝑥 ]10 − ∫ 𝑒 𝑥 𝑑𝑥
0
0
= [𝑥 𝑒 𝑥 ]10 − [𝑒 𝑥 ]10
= 𝑒−𝑒+1= 1
EXAMPLE 10
𝑒
∫ 𝑥 3 ln 𝑥 𝑑𝑥
1
Solution:
Since we cannot directly integrate ln x, therefore u will be ln x. Since the integral is a definite integral,
the final answer must not be in terms a function.
9
Let
𝑑𝑣 = 𝑥 3 𝑑𝑥
𝑢 = ln 𝑥
1
1
𝑣 = ∫ 𝑥 3 𝑑𝑥 = 4 𝑥 4
𝑑𝑢 = 𝑥 𝑑𝑥
By applying the formula,
∫ 𝑢 𝑑𝑣 = 𝑢𝑣 − ∫ 𝑣 𝑑𝑢
𝑒
𝑒
𝑒
1
1
1
∫ 𝑥 𝑙𝑛 𝑥 𝑑𝑥 = [ 𝑥 4 𝑙𝑛 𝑥] − ∫ ( 𝑥 4 ) ( ) 𝑑𝑥
4
4
𝑥
1
3
1
1
𝑒
𝑒4 1
=
− ∫ 𝑥 3 𝑑𝑥
4
4
1
𝑒
𝑒4 1 𝑥4
=
− [ ]
4
4 4 1
=
1
(3𝑒 4 − 1) (Verify this)
16
By the end of this sub-topic you should know:
1.
2.
3.
4.
The formula for integrating by parts.
Have some ideas on how the formula was formulated and why (theory part!)
How to choose strategically the u and dv.
Not all integral can straight away apply the technique of integration by parts, need to use basic
substitutions.
5. There are special cases of integral by parts.
10
EXERCISES IN CLASS
Evaluate the following integral using integration by parts. (Some of the integrals need basic substitution
before integrating)
1.
3.
5.
2.
∫ 𝑥 3 ln 3𝑥 𝑑𝑥
∫ 𝑥 2 𝑒 4𝑥
4.
∫ 𝑒 2𝑥 cos 3𝑥 𝑑𝑥
ln 𝑡
𝑑𝑡
𝑡
6.
Show that :
8.
∫ 3𝑥 cos 2𝑥 𝑑𝑥
∫
𝑒
∫ ln 𝑥 𝑑𝑥
1
7.
1
3
∫ 𝑥 2 𝑒 −3𝑥 𝑑𝑥 =
0
9.
∫ 𝑥 5 ln(𝑥 6 ) 𝑑𝑥
Show that :
4
2𝑒 − 5
27𝑒
∫ 𝑥 ln 𝑥 𝑑𝑥 = 14 ln 2 − 3
2
10.
TUTORIAL
X siap gi…
11
∫ 𝑡 sin−1 (𝑡 2 − 1) 𝑑𝑡