Integration by parts.
Integration by parts.
We can integrate equation using integration by parts using the formula;
= uv- 
Let us look at the guidelines for selecting u and dv-There are always exceptions, but these are helpful.
We use the acronym L-I-A-T-E as a guide when selecting u to be the function that comes in this list first.
L: Logarithm function.
I: Inverse trig function.
A: Algebraic function.
T: Trigonometric function.
E: Exponential function.
Sample problem 1.
Integrate 
Solution.
From the equation, In x is a logarithmic function and x is an algebraic function. Therefore, we let;
u=In x and dv = x (Since L comes before A in LIATE)
du = 
dx and v= 
= 
x2
Therefore,
= uv-
= (In x)x2 - 
x2 dx
= (In x)x2 - (
x3) + C
= (In x)x2 - 
x3 + C …………………………………………………………Answer.
Sample problem 2.
Solve 
Solution
From the equation, In x is a logarithmic function and x is an algebraic function. Therefore, we let;
u=
and dv = sin x (Since L comes before T in LIATE)
du = (-tan x) dx and v= 
= 
Therefore,
= uv-
= (In cos x) (-cos x) - 
(-tan x) dx
= -cos x (In cos x) - 
dx
= -cos x (In cos x) -
= -cos x(In cos x) + cos x + C ……………………………………Answer.
Sample problem 3.
Integrate 
dx
Solution.
At glance it appears that integration by parts does not apply, but let;
u = 
(Inverse Trig function) and dv= 
(Algebraic function)
du= 
x2)-1dx whereas v = 
= x
Therefore,
dx = uv-
= (
- 
x2)-1dx
= x (
– (
) 
(1 - x2)1/2 (-2x) + C
= x ( + () (1 - x2)1/2 (2) + C
= x ( + (1 - x2)1/2 + C ……………………………………………………Answer.
Alternative general guidelines for selecting u and dv:
Let u be that section of the integrand whose derivative du is a “simpler” function than u itself.
Let dv be the most complicated section of the integrand that can be integrated easily.
Sample problem 4.
Solve x3 (4 –x2)1/2 dx
Solution.
The LIATE rule of thumb isn’t applicable here since both x3 and (4 –x2)1/2 are algebraic functions. We apply part 2 of the alternative guidelines above and find that x (4 –x2)1/2 is the most intricate section of the integrand that can be easily integrated. Therefore;
dv = x (4 –x2)1/2 and u = x2 (the remaining factor in the integrand)
This means that du = 2xdx and
v= (4 –x2)1/2 dx =- 
(4 –x2)1/2 dx
= (-) (
) (4 –x2)3/2 = - (4 –x2)3/2
We can now solve our problem.
x3 (4 –x2)1/2 dx = uv-
= (x2) (- (4 –x2)3/2 ) - 
(4 –x2)3/2 (2x) dx
= -x2(4 –x2)3/2 - (4 –x2)5/2 (
) + C
= -x2(4 –x2)3/2 - 
(4 –x2)5/2 + C …………………………………Answer.
Using repeated application of integration by parts.
Occasionally integration by parts can be repeated in order to get the final solution, but caution should be taken by not substituting the choices for u and dv in the successive applications.
Sample problem 5.
Integrate x2 sin x dx
Solution.
From the equation, sin x is a trigonometric function and x2 is an algebraic function. Therefore, we let;
u=x2 and dv =sin xdx (Since A comes before T in LIATE)
du = 2x dx and v= = -cos x
Therefore,
x2 sin x dx = uv-
= x2(-cos x) - 
= -x2 cos x +2 
Applying integration by parts for the second time to solve 
the second part of the integral above) and repeating the same procedure of choosing u and dv
u=x (Algebraic function) and dv = cos x (trigonometric function)
Therefore; du= dx and v=
= sin x
x2 sin x dx = -x2 cos x +2(uv-)
= -x2 cos x +2(x sin x
)
= -x2 cos x +2(x sin x + cos x + C)
= -x2 cos x +2x sin x + 2cos x + C ………………………………………Answer.
It is important to note that the constant multiple, 2 from the original integral should not be ignored.
Sample problem 6.
Solve ex cos x dx.
Solution
From the equation, ex is an exponential function and cos x is a trig-function. Therefore, we let;
u=
and dv = ex (Since T comes before E in LIATE)
du = -sin x dx and v=ex dx = ex
Therefore,
ex cos x dx.
= uv-
Cos x ex - ex (-sin x) dx.
Cos x ex + ex sin x dx.
Applying integration by parts for the second time to solveCos x ex + ex sin x dx, 
the second part of the integral above) and repeating the same procedure of choosing u and dv
u= sin x (Trigonometric function) and dv = ex (exponential function)
Therefore; du= cos x dx and v= ex dx= ex
ex cos x dx.= Cos x ex + (uv-)
ex cos x dx.= Cos x ex + sin x - ex cos x dx
The original integral appears on the R.H.S of the equation; therefore we collect like terms by moving it to the L.H.S as follows;
ex cos x dx + ex cos x dx= ex Cos x + sin x
2ex cos x dx.= ex Cos x + sin x + C
ex cos x dx = (ex Cos x + sin x) + C …………….…………………………………Answer.
Exercise.
(Clue=x & dv= sin x cos x )
= sin x cos x
- x3 (9 – x2)1/2 dx (Clue; u=x2 & dv= x (9 – x2)1/2 dx )
3. 
(Clue=cos x-1 & dv=dx)
