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Methods of Integration
William Gunther
June 15, 2011
In this we will go over some of the techniques of integration, and when to apply them.
1 Simple Rules
So, remember that integration is the inverse operation to differentation. Thuse we get a few rules for free:
Sum/Difference R(f(x)±g(x))dx=R f(x)dx±R g(x)dx
Scalar Multiplication R cf(x) dx = c·R f(x) dx for c ∈ R
R xn+1
Product Rule xn dx = +Cfor n6=−1
n+1
The above allows us to integrate any polynomials and roots. The only think we don’t yet know how to
integrate is R 1 dx. Luckily, we know d ln(x) = 1. From this, and other knowledge we know about
x dx x
derivatives, we know:
Trig
R sin(x) dx = −cos(x)+C
R cos(x) dx = sin(x) +C
R sec2(x) dx = tan(x)+C
R sec(x)tan(x) dx = sec(x)+C
Exponentials
R x x
e dx=e +C
R 1 dx = ln|x|+C.
x
!!EXAMPLES!!
2 u-substitution
Notice, if f(x) and g(x) are functions, then the chain rule says
d (f(g(x))) = f′(g(x))·g′(x)
dx
So, we know: Z
f′(g(x)) · g′(x) dx = f(g(x))
Writing this out in a better way, we get let u = g(x). Then du = g′(x) dx, meaning we can trade a
g′(x) dx for a du and substitute u for g(x) in the integral. The goal is to eliminate all occurrences of x in
the integral, and then your entire integral is in terms of u, and is simplier.
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Example 1. Let us solve the integral Z
sin(2x) dx
We do this by doing the substitution u = 2x. Then du = 2 dx. Thus we can trade a 2 dx for a du. So we
write the integral in the following way:
Z sin(2x) dx = 1 Z sin(2x)(2 dx)
2
Then: Z Z
1 sin(2x)(2 dx) = 1 sin(u) du
2 2
Doing the integration: Z
1 sin(u) du = 1(−cos(u))+C
2 2
As the problem was given in terms of x, we want the answer in terms of x. So we substitute 2x for u.
1(−cos(u))+C =−cos(2x) +C
2 2
Wedothe following integrals with less exposition:
Example 2. Z
2
xcos(x ) dx
Set u = x2. Then du = 2x dx.
Z 2 1 Z 2
xcos(x ) dx = 2 cos(x )2x dx
= 1 Z cos(u) du
2
= 1(sin(u))+C
2
sin(x2)
= 2 +C
Example 3. Z
cos(ln(x)) dx
x
Set u = ln(x). Then du = 1 dx.
x
Z cos(ln(x)) dx = Z cos(ln(x))1 dx
x Z x
= cos(u)du
=sin(u)+C
=sin(ln(x))+C
Example 4. Z
3cos(x)esin(x) dx
Let u = sin(x). Then du = cos(x) dx.
Z sin(x) Z sin(x)
3cos(x)e dx = 3·Z cos(x)e dx
u
=3· e du
=3·eu+C
=3esin(x) +C
2
Example 5. Z
10
x(x+5) dx
Here, we can solve the integral by expanding. But, expanding the 10th power is rather annoying. So instead:
Let u = x+5. Then we get x = u−5, and du = dx. All we have done is a linear transformation. Note,
in general we can not solve for x when we do a substitution. When the substitution is linear we can.
Z x(x+5)10 dx = Z (u−5)(u)10 du
Z 11 10
= (u −5u )du
12 11
= u −5u +C
12 11
(x+5)12 5(x+5)11
= 12 − 11 +C
3 Integration by Parts
Recall the product rule:
d [f(x)g(x)] = f′(x)g(x)+f(x)g′(x)
dx
Moving things around, we see
f′(x)g(x) = d [f(x)g(x)]−f(x)g′(x)
dx
Integrating both sides, we see
Z f′(x)g(x) dx = f(x)g(x)−Z f(x)g′(x)dx
Renaming v = f(x) and u = g(x) we have dv = f′(x) dx and du = g′(x) dx and our formula becomes
Z udv =uv−Z vdu
Here, we seperate our integral into two parts: one part we differentiate, and the other we integrate. Then
we apply the formula, and get a new integral with these new parts (the derivative of the one part and the
integral of the other).
As a strategy, we tend to choose our u (the part we differentiate) so that the new integral is easier to
integrate. We also need to take care that the dv (the part we integrate) can actually be integrated by us.
Example 6. Z
x
x·e dx
Here, we see that when we take the derivate of x it vanishes completely making our next integral simplier.
Z x
x·ex dx u=x dv = e dx
du = dx v = ex
x Z x
=xe − e dx
x x
=xe −e +C
Asaheuristic (rule of thumb) we choose logarithms and inverse trigonometric functions to be our u before
any others since their integrals are hard to calculate and complicated. After those, we like polynomials (or
really anything algebraic) as those derivatives often get simplier. We rarely want to choose exponentials to
be our u since integrating an exponential is virtually the same as deriving it.
Unless we deviate from this heuristic, the u shall be chosen without exposition:
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Example 7. Z
ln(x) dx
Wedothis by parts:
Z ln(x) dx u=ln(x) dv = dx
du = 1 dx v = x
Z x
=xln(x)− 1xdx
Z x
=xln(x)− dx
=xln(x)−x+C
Example 8. Z
1
2 −x
(x +1)e dx
0
Wedothis by parts:
Z 1 2 −x
(x2 +1)e−x dx u=x +1 dv=e dx
0 Z du = 2x dx v = −e−x
2 −x −x
=(x +1)(−e )− (−e )(2x)dx
2 −x Z −x u=x dv = e−x dx
=−(x +1)e +2 xe dx du = dx v = −e−x
2 −x −x Z −x
=−(x +1)e +2(−xe − (−e )dx)
2 −x −x Z −x
=−(x +1)e −2xe +2 e dx
2 −x −x −x
=−(x +1)e −2xe −2e +C
Sometimes, we can do a nice subsitution before finishing the problem using parts:
Example 9. Z
2
3 x
x ·3 dx
b bln(a)
First, we re-write it to have e as the exponential base. Note that in general a = e So
Z 3 x2 Z 3 x2ln(3)
x ·3 dx = x ·e dx
Let u = x2. Then du = 2x dx. So 1 du = x dx.
2
Z 2 Z 2
3 x ln(3) 2 (x )ln(3)
x ·e dx = Z (x )·e (x dx)
uln(3)
= u·e du
Now, we do parts.
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