Integration by Parts
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What Kind of Problems Can Be Applied
- This technique can be applied to a wide
variety of functions and is particularly
useful for integrands involving products of
algebraic and transcendental functions.
General Theorem
If u and v are function of x and have
continuous derivatives, then
∫u dv = uv - ∫v du
Example 1
Evaluate ∫sec3x dx
Set u = sec x and dv = sec2x dx
So du = secxtanx dx v = tan x
then ∫sec3x dx
= secxtanx - ∫secxtan2xdx
= secxtanx - ∫secx(sec2x-1)dx
= secxtanx - ∫sec3xdx + ∫secxdx
So, 2∫sec3x dx = secxtanx + ln|secx + tanx| + C
∫sec3x dx = (1/2)(secxtanx + ln|secx + tanx| + C)
Example 2
Evaluate ∫arcsinx dx
Let u = arcsinx and
So du = 1/√(1-x2)dx
dv = dx
v=x
then ∫arcsinx dx
= xarcsinx - ∫x/√(1–x2) dx
= xarcsinx + √(1-x2) + C
Guidelines for
Integration by Parts
1. Let dv be the most complicated portion of
the integrand that fits a basic integration
rule. Then u will be the remaining factor(s)
of the integrand.
2. Let u be the portion of the integrand
whose derivative is a function simpler than
u. Then dv will be the remaining factor(s)
of the integrand.
Common Integrals using Integration by Parts
1. For integrals of the form
∫xn eax dx
∫xn sin(ax) dx
∫xn cos(ax) dx
Let u = xn
and dv = eax dx, sin(ax) dx, cos(ax) dx
2. For integrals of the form
∫xn lnx dx
∫xn arcsin(ax) dx
∫xn arctan(ax) dx
Let u = lnx, arcsin(ax), or arctan(ax)
and dv = xn dx
3. For integrals of the form
∫eax sin(bx) dx
∫eax cos(bx) dx
Let u = sin(bx) or cos(bx)
and dv = eax dx
Example 3
Evaluate ∫lnx dx
Let u = lnx and dv = dx
So du = 1/x dx
v=x
Then ∫lnx dx
= xlnx - ∫x(1/x)dx
= xlnx – x + C
Example 4
Evaluate ∫exsinx dx
Let u = sinx and
So du = cosx dx
dv = ex dx
v = ex
Then ∫ex sinx dx
= exsinx - ∫excosx dx
= exsinx - (excosx - ∫ex(-sinx) dx)
= exsinx - excosx - ∫exsinx dx
So, 2∫exsinx dx = exsinx – excosx
∫exsinx dx = (1/2)(exsinx - excosx) + C
Example 5
Evaluate ∫x2sin(4x) dx
Let u = x2 and
So du = 2x dx
dv = sin(4x) dx
v = -cos(4x)/4
Then ∫x2sin(4x) dx
= -(x2cos(4x))/4 - ∫(-cos(4x)/4)(2x) dx
= -(x2cos(4x))/4 + (1/2)∫(xcos(4x) dx
= -(x2cos(4x))/4 + (1/2)(xsin(4x)/4 - ∫sin(4x)/4 dx)
= -(x2cos(4x))/4 + (1/2)(xsin(4x)/4 + cos(4x)/16) + C
Tabular Method
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