MATH 126 midterm 1 preparation guide
In midterm 1 there will be covered sections 5.6, 5.8, 6.1, 6.2, 6.3:
ˆ Section 5.6, inverse trigonometric functions: definition, domain and range of arcsin, arccos, arctan
functions. Derivatives of inverse trigonometric functions. Typical exercises:
1. Find tan(arccos( 21 )), arcsin(cos(4)).
2. Compute the derivative of f (x) = sin−1 (ln(x + 1))
ˆ Section 5.8, l’Hospital rule, indeterminate forms: indeterminant fractions, indeterminant products,
indeterminant differences, indeterminant powers. Typical exercises:
lim
x→0
tan(x)
1
, lim (1 + 5x)1/x , lim (1 + 5x)1/x , lim x3 ln(1 + ),
x→∞
x→∞
x + sin(x/2) x→0
x
ˆ Section 6.1, integration by parts.
R
R
R
R
2
Typical exercises: x3 e−x dx, arctan(3x + 5)dx, arccos(x), e5x cos(x)dx
ˆ Section 6.2, trigonometric integrals, integrating functions of the form sinn (x) cosm (x), using of double
angle √
formula to√reduce powers
of sin and cos, trigonometric substitutions to deal with radicals of the
√
form x2 + a2 , x2 − a2 , a2 − x2 .
R dx
R
R
Typical exercises: √5−x
dx, (x22x+1
dx, x3 (2 + x2 )5/2 dx
2
−4)5/2
ˆ Section 6.3, partial fractions:
computing the integrals of any rational function in two steps: first,
R
dx
where deg p < deg q; second, decomposing the denominator q(x) as
reducing to the case p(x)
q(x)
product of irreducible polynomials of degree
R one1and two, reducing to the partial fractions. Recall that
dealing with partial fractions of the form (x2 +a
2 )n dx we need to use substitution x = a tan(t) when
n > 1.
R 2
R x2 −x+5
Typical exercises: xx2+3x+1
+x−2 dx,
(x2 +x+4)2 dx.
Note that in many cases it is necessary to use a combination of methods we learned to get the answer, for
example in
Z
ln(1 + x2 )dx.
The following method, called tangent half-angle substitution, not mentioned in the textbook. It allows to
reduce computation of integral of any rational trigonometric function (i.e. rational function in sin and cos)
to the case of ordinary rational function:
1−t2
2dt
2t
If t = tan(x/2), then sin(x) = 1+t
2 , cos(x) = 1+t2 , and dx = 1+t2 .
For example,
Z
Z
Z
Z
1
1 + t2 2dt
2dt
1
1
dx =
=
=
+
dt = − ln(1 − t) + ln(1 + t) + C =
cos(x)
1 − t2 1 + t2
1 − t2
1−t 1+t
1 + tan(x/2)
= ln
+C
1 − tan(x/2)
1