Download 3450:335 Ordinary Differential Equations Review of Integration and

3450:335 Ordinary Differential Equations
Review of Integration and Trigonometry
Here is a review of the integration and trigonometric skills you need in this course.
Integration by substitution
Z
Z
p
e2x dx =
2x3 1 + x2 dx =
1 2x
e + c u = 2x, du = 2dx
Z2
Z
(u − 1)u1/2 du =
u3/2 − u1/2 du
u = 1 + x2 , du = 2xdx, x2 = u − 1
Integration by parts
There
Z are 3 classes of integrals that
Z can be evaluated using IBP:
A)
(polynomial)(sin or cos) dx,
polynomial
one degree at a time.
Z
B)
(polynomial)(logarithm) dx,
(polynomial)(exp) dx. Let (u = polynomial) to kill off the
Z
arctrig dx. Let u = logarithm or u = arctrig.
Z
C)
(exp)(sin or cos) dx. IBP twice, using u = trig both times, or u = exp both times. Called
FOLDING.
Basic Trigonometric Skills
A) Definitions of the 6 trig functions in terms of right triangles and in terms of sines and cosines.
B) Pythagorean identities, 3 flavors. sin2 x + cos2 x = 1, 1 + cot2 x = csc2 x, tan2 x + 1 = sec2 x.
C) Half angle identities. sin2 x = 21 (1 − cos(2x)), cos2 x = 12 (1 + cos(2x)).
D) Double angle identities. 2 sin x cos x = sin(2x), cos2 x − sin2 x = cos(2x), 2 cos2 x − 1 =
cos(2x), 1 − 2 sin2 x = cos(2x).
E) Sum and difference identities. sin(x ± y) = sin x cos y ± sin y cos x, cos(x ± y) = cos x cos y ∓
sin x sin y.
F) Product identities. sin A cos B = 12 (sin(A − B) + sin(A + B)), sin A sin B = 21 (cos(A − B) −
cos(A + B)), cos A cos B = 12 (cos(A −
B) + cos(A + B)). R
R
R
G) Elementary
antiderivatives. R sin x dx = cos x + c, cos x dx = − sin x + c, sec2 x dx =
R
tan x + c, csc2 x dx = − cot x + c, sec x tan x
Z dx = sec x +Z c.
Z
1
sin x
H) Antiderivatives using logarithmic form. tan x dx =
dx = −
du = − ln | cos x|+
cos x
u
Z
Z
Z
1
sec x + tan x
c, sec x dx = sec x
dx =
du = ln | sec x + tan x| + c.
sec x + tan x
u
Trigonometric
Integrals
Z
A)
sinm x cosn x dx. If n or m is odd, pull off the odd power of sin or cos to use in du, let
u equal the other trig function, and apply Pythagorean identities if needed. If both n and m are
odd, you can let u equal sin x or cos x. If n and m are both even, apply half angle identities and
Pythagorean identities.
1
Examples:
Z
Z
3
cos x dx =
Z
sin3 x cos3 x dx =
Z
Z
2
(1 − sin x)(cos xdx) =
Z
Z
(u3 )(1 − u2 )(du)
1
(1 − cos(2x)) dx
2
sin x dx =
B)
(1 − u2 )du
(sin3 x)(1 − sin2 x)(cos xdx) =
Z
2
Z
tanm x secn x dx. If n is even, pull off a sec2 x to use in du, let u = tan x and apply
Pythagorean identity if needed. If m is odd, pull off a sec x tan x to use in du, let u = sec x, apply
Pythagorean identity if needed. If both n is even and m is odd, you can do either of the above. If
n is odd and m is even, God help you. Use Pythagorean identities to reduce to nothing but powers
of sec x. Integrals of sec x and sec3 x are standard, and can be looked up in your old Calculus text
(you didn’t sell it back to the Bookstore, did you?). Higher odd powers of sec x require careful use
of IBP. Even powers can be handled by the second sentence in this paragraph.
Examples:
Z
2
4
Z
tan x sec x dx =
Z
Z
2
2
2
tan x(tan x − 1)(sec x dx) =
tan x sec20 x dx =
Z
sec19 x(sec x tan x dx) =
tan2 x sec x dx =
Z
(sec2 x − 1) sec x dx =
Z
Z
Z
u2 (u2 − 1) du
u19 du
sec3 x − sec x dx
Trigonometric Substitution
√
√
√
Look for expressions of the form a2 − x2 , x2 − a2 or x2 + a2 , or one of these to a higher
power (for example, (x2 − 4)2 is the square root raised to the fourth power). If there is a minus
under the square root, the positive term is the square of the hypotenuse and the negative term is
the square of one of the legs; the square root is the other leg. If there is a plus under the square
root, then the square root is the hypotenuse, and the terms inside are the squares of the legs. Use
the triangle to make a substitution, which converts the integral to a trig integral (see above).
Examples:
Z
√
1
√
1)
dx. The hypotenuse is 4 and the sides are x and 16 − x2 . Let x = 4 sin θ,
x2 16 − x2
Z
Z
1
1
p
4 cos θ dθ =
1 sin2 θ dθ =
so dx = 4 cos θ. The integral becomes
2
2
16
16
sin
θ
16
−
16
sin
θ
√
1 16 − x2
1
− cot θ + c = −
+ c.
x
16
16
√
x3
dx. The hypotenuse is 9 + x2 and the legs are x and 3. Let x = 3 tan θ, so
2) √
9 + x2
Z
Z
27 tan3 θ
sin3 θ
2
2
dx = 3 sec θdθ. The integral becomes
3 sec θ dθ = 27
dθ. Write the numerator
3 sec θ
cos4 θ Z
as sin2 θ(sin θ dθ) = (1 − u2 )du for u = cos θ. The integral becomes 27 (1 − u2 )u−4 (−du) =
p
√
1
1
1
27 − + 3 = −9 9 + x2 + (9 + x2 )3/2 . Note that u = cos θ = 3/ 9 + x2 . Oh, and
u 3u
3
don’t forget the constant of integration.
2