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Integration Methods
Elyse Yeager
Math 105
Wehave learned several fancy methods of integration. Although occasionally one method might be the
“obvious” choice, in general it is quite difficult to decide how to proceed when faced with an antiderivative.
The only thing you can do is practice, practice, practice.
Substitution
This is doing the chain rule in reverse. If you see some function whose derivative shows up multiplied, you
might want to try substitution.
Easy example: R(4x+2)cos(2x2 +2x)dx, using u = 2x2 +2x.
Parts
This is doing the product rule in reverse. If you see the product of two functions, and you would like to
swap one with its derivative and the other with its antiderivative, you might want to try integration by
parts. R
x x
Easy example: xe dx, u = x, dv = e dx.
Trig Integrals
R n m R n m
Wecan handle integrals of the type sin xcos xdx and tan xsec xdx using the rules in Chapter 7.3.
They involve substitutions and trig identities.
2 2 2 2 2 1−cos2x 2 1+cos2x
Identities to remember: sin x+cos x = 1, tan x+1 = sec x, sin x = 2 , and cos x = 2 .
Trig Substitution
Mostly, this comes up when you have a quadratic equation underneath a square root. Using a substitution
allows you to take advantage of trig identities to turn the thing under the square root into a perfect square.
2 2 2 2
You will have to remember the identities sin x+cos x = 1 and tan x+1 = sec x. After you get rid of
the square root, you still might have to use methods from trig integrals to evaluate, perhaps using another
substitution. To get everything in terms of x, you might have to draw a triangle.
R √ 2
Easy Example: 1−x dx. Use x = sinθ.
Partial Fractions
Partial fractions is a method for evaluating the integral of a rational function (a fraction whose numerator
and denominator are both polynomials). Partial fractions ONLY works on rational functions, and it gives
you an integral of MORE rational functions, so it will usually lead you down a road to substitution. For
this reason, if you see a rational function that you can already integrate using substitution, do that; if you
see a rational function that you can’t integrate using substitution, try partial fractions to turn it into a
friendlier form. R
Arctangents often come up here: remember 1 dx = arctanx+C. Also, you’ll need to factor
2
1+x
polynomials. A guide to the decomposition is given on Page 548 of your text.
Easy example: R x+1 2dx
(3x+5)
1
Problems
Question 1. Z
π
4 5
sin xcos xdx
0
Question 2. Z p
3−5x2dx
Question 3. Z
2
2
x lnxdx
1
Question 4. Z √
e ln x
x dx
1
Question 5. Z
0.2 tanx
ln(cosx)dx
0.1
Question 6. Z
−2 dx
2
3x +4x+1
Question 7. Z √
2
x −2dx
x2
Question 8. Z
π/4
4 5
sec xtan xdx
0
Question 9. Z
2
3x +4x+6dx
(x+1)3
Question 10. Z
2
(3x) arcsinxdx
Question 11. Z
1 dx
2
x +x+1
Solutions
Question 1
Z π 4 5
sin xcos xdx
0
This is a product of sine and cosine. Since cosine has an odd power, we can reserve one cosine for the
derivative of sine in our substitution, change the rest of the cosines to sine, and let u = sinx.
Z π 4 5 Z π 4 2 2
sin xcos xdx = sin x(cos x) cosxdx
0 0
Z π 4 2 2
= 0 sin x(1−sin x) cosxdx
Z x=π 4 2 2
= x=0 u (1−u ) du
Z x=π 4 2 4
= x=0 u (1−2u +u )du
Z x=π 4 6 8
= x=0 (u −2u +u )du
x=π
1 5 2 7 1 9
= u − u + u
5 7 9
x=0
π
1 5 2 7 1 9
= sin x− sin x+ sin x
5 7 9
0
1 5 2 7 1 9 1 5 2 7 1 9
= 5sin π− 7sin π+ 9sin π − 5sin 0− 7sin 0+ 9sin 0
=0
Question 2
Z p3−5x2dx
Wenotice that there is a quadratic equation under the square root. If that equation were a perfect square,
we could get rid of the square root: so we’ll mold it into a perfect square using a trig substitution.
Our candidates will use one of the following identities:
2 2 2 2 2 2
1−sin θ =cos θ tan θ+1=sec θ sec θ −1 = tan θ
2
We’ll be substitution x =(something), so we notice that 5 −5x has the general form of
2
(constant)−(function), as does 1−sin θ. In order to get the constant right, we multiply through by three:
2 2
3−3sin θ =3cos θ
2 2
Our goal is to get 3 − 5x = 3−3sin θ; so we solve for x and decide on the substitution
x=r3sinθ, dx = r3cosθdθ
5 5
Now we evaluate our integral.
v !
Z Z u r 2r
p u 3 3
2 t
3−5x dx= 3−5 5 sinθ 5 cosθdθ
Z p 2 p
=Z 3−3sin θ 3/5cosθdθ
√ 2 p
= 3cos θ 3/5cosθdθ
Z √ p
= 3cosθ 3/5cosθdθ
3 Z 2
= √5 cos θdθ
3 Z 1+cos2θ
= √5 2 dθ
3 Z
= √ (1+cos2θ)dθ
2 5
3 1
= √ θ+2sin(2θ) +C
2 5
3
= √ [θ+sinθcosθ]+C
2 5
p p p
From our substitution x = 3/5sinθ, we glean sinθ = x 5/3, and θ = arcsin x 5/3 . To figure out
√ √
x 5
cosθ, we draw a triangle. We let θ be an angle, and since sinθ = √3 , we let the hypotenuse be 3 and the
√ √ 2
side opposite θ be x 5. By Pythagorus, the missing side (adjacent to θ) has measure 3−3x .
√
√ 3
x 5
θ
√ 2
3−5x
adj √ 2
3−5x
Therefore, cosθ = hyp = √ . So our integral is:
3
" √ √ p √ 2#
3 3 3−5x
√ [θ+sinθcosθ]+C = √ arcsin(x 5/ 3)+x 5/3· √ +C
2 5 2 5 3
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