Download Derivatives and Integrals of Trigonometric and Inverse

Calculus 2
Lia Vas
Derivatives and Integrals of Trigonometric and Inverse
Trigonometric Functions
Trigonometric Functions.
If y = sin x, then y 0 = cos x and
if y = cos x, then y 0 = − sin x.
R
R sin x dx = − cos x + c and
cos x dx = sin x + c.
Thus,
The derivatives and integrals of the remaining trigonometric functions can be obtained by expressx
sin x
, cot x = cos
,
ing these functions in terms of sine or cosine using the following identities: tan x = cos
x
sin x
sec x = cos1 x , csc x = sin1 x .
Example 1. Find derivative of tan x. Simplify your answer.
Solution. Using the formula tan x =
cos2 x+sinx x
= cos12 x or sec2 x.
cos2 x
sin x
cos x
and the quotient rule, obtain
d tan x
dx
=
cos x cos x−(− sin x) sin x
cos2 x
≤x≤
Inverse Trigonometric Functions. The function sin x passes horizontal line test for −π
2
−1
π
so
it
has
an
inverse.
The
inverse
function
is
denoted
by
sin
x
or
arcsin
x.
Since
the
range
of
sin x
2
−1
−π π
on [ 2 , 2 ] is [-1,1,], the interval [-1,1] is the domain of sin x. We also have the following cancellation
rule.
π
−π
≤x≤
sin(sin−1 x) = x for − 1 ≤ x ≤ 1 and sin−1 (sin x) = x for
2
2
Similarly, one obtains cos−1 x on [−1, 1] is the inverse of cos x for 0 ≤ x ≤ π and the analogous
< x < π2 . The
cancellation rule holds. The function tan−1 x on (−∞, ∞) has the inverse tan x for −π
2
inverses of other trigonometric functions can be obtained similarly.
Careful: when using notation sin−1 x do not mix this function up with (sin x)−1 =
1
.
sin x
When solving an equation of the form sin x = a for x where a is a number in [−1, 1], the second
cancellation formula implies that x = sin−1 a is one solution of this equation and that this value will
be in interval [ −π
, π ]. In many cases, we need more than this one solution, in particular, the second
2 2
solution in interval [0, 2π].
Using the trigonometric circle (or the graph of
sine function) we can see that if
x1 = sin−1 a,
the second solution can be obtained as
x2 = π − x1 = π − sin−1 a.
1
=
Similarly, when solving cos x = a, in interval
[0, 2π] the first solution can be found as
x1 = cos−1 a
and the second solution can be obtained as
x2 = −x1 = − cos−1 a.
For the equation tan x = a, on [0, 2π], the first
solution can be found as
x1 = tan−1 a
and the second solution can be obtained as
x2 = π + x1 = π + tan−1 a.
Derivatives of the Inverse Trigonometric Functions.
The formula for the derivative of y = sin−1 x can be obtained using the fact that the derivative
of the inverse function y = f −1 (x) is the reciprocal of the derivative x = f (y).
y = sin−1 x ⇒ x = sin y ⇒ x0 = cos y ⇒ y 0 =
1
1
1
.
=
=
x0
cos y
cos(sin−1 x)
To be able to simplify this last expression, one needs to represent cos y in terms of sin y. This can be
done using the trigonometric identity
q
q
√
2
2
−1
2
sin y + cos y = 1 ⇒ cos y = 1 − sin y ⇒ cos(sin x) = 1 − (sin(sin−1 x))2 = 1 − x2 .
Thus, we obtain the formula for the derivative of y = sin−1 x to be
y0 = √
1
1 − x2
Similarly, one obtains the following formulas.
d
1
(tan−1 x) = 2
dx
x +1
d
1
(sec−1 x) = √
dx
x x2 − 1
−1
Differentiating the trigonometric identities sin x + cos−1 x = π2 , tan−1 x + cot−1 x = π2 , and
−1
sec x + csc−1 x = π2 , we obtain that the derivatives of cos−1 x, cot−1 x, and csc−1 x are negative of
the derivatives of sin−1 x, tan−1 x, and sec−1 x respectively.
Integrals producing inverse trigonometric functions. The above formulas for the the
derivatives imply the following formulas for the integrals.
Z
1
√
dx = sin−1 x + c
2
1−x
Z
Z
1
1
−1
√
dx
=
tan
x
+
c
dx = sec−1 x + c
2
x2 + 1
x x −1
2
Practice Problems:
1. Find all solutions of the following equations on interval [0, 2π].
(a) sin x =
2
5
(b) cos2 x =
1
4
(c) 2 tan x + 3 = 9
(d) 2 cos2 x + cos x − 1 = 0
(e) sin x cos x = sin x
(f) sin x = cos x
2. Find the derivatives of the following functions.
(a) y = cot x2
(b) y = x2 cos x2
(c) y = sin(3x + 2) cos(2x − 3)
(d) y = sin−1 (2x)
(e) y = sin(ax) cos(bx) where a and b are arbitrary constants.
(f) y = sin−1 (x4 )
(g) y = x2 cos−1 x
(h) y = tan−1 (ex )
(i) y = etan
−1
3. Evaluate the following integrals.
R
R
(a) cos(3x + 1) dx
(b) x sin x2 dx
R1 1
R
1
(d) 0 √1−x
(e) √1−4x
2 dx
2 dx
R x+1
R 1
(h) x2 +4 dx
(g) x2 +4 dx
R x+6
(j) (Extra credit level) x2 +4x+13 dx
x
R
(c) (9 + 2 sin πx
) dx
5
R 1
(f) 4x2 +1 dx
R
(i) xx+3
2 +9 dx
4. Find the area of the region between y = sin x and y = cos x for x in [0, 2π].
Solutions.
1. (a) sin x = 52 ⇒ x = sin−1
and 156.42 degrees.
2
5
and x = π − sin−1
2
5
⇒ x ≈ .411 and x ≈ 2.73 radians or 23.57
(b) cos2 x = 14 ⇒ cos x = ± 12 ⇒ x = ± cos−1 12 , x = ± cos−1 −1
⇒ x = ± π3 , x = ± 2π
.
2
3
4π
5π
,
and
or
60,
Converted to values in the interval [0, 2π], the four solutions are π3 , 2π
3
3
3
120, 240, and 300 degrees.
(c) 2 tan x + 3 = 9 ⇒ 2 tan x = 6 ⇒ tan x = 3 ⇒ x = tan−1 (3) and x = π + tan−1 (3) ⇒
x ≈ 1.25 and x ≈ 4.39 radians or 71.56 and 251.57 degrees.
(d) 2 cos2 x + cos x − 1 = 0 ⇒ (2 cos x − 1)(cos x + 1) = 0 ⇒ cos x = 12 , cos x = −1 ⇒
x = ± cos−1 21 = ± π3 , x = ± cos−1 (−1) = ±π. Converted to values in the interval [0, 2π],
the solutions are π3 , π, and 5π
or 60, 180, and 300 degrees.
3
(e) sin x cos x = sin x ⇒ sin x cos x − sin x = sin x(cos x − 1) = 0 ⇒ sin x = 0, cos x = 1. The
first equation has solutions 0 and π and the second just 0. Thus 0 and π are the solutions.
(f) One way to solve this equation is to divide by cos x so that it reduces to tan x = 1. Thus
x = tan−1 (1) = π4 and x = π + tan−1 (1) = 5π
. Another way to do it is to convert cos to
4
p
sine (or vice versa). In this case the equation reduces to sin x = 1 − sin2 x ⇒ sin2 x =
1 − sin2 x ⇒ sin2 x = 21 ⇒ sin x = ± √12 and be careful about the extraneous roots 3π
and
4
7π
.
4
3
2
x
2. (a) Representing the function as y = cot x2 = cos
and using the quotient rule with f (x) =
sin x2
2
2
0
cos x and g(x) = sin x and the chain for f (x) = − sin x2 (2x) and g 0 (x) = cos x2 (2x)
2 2
2
2 x2 )
x2 −cos x2 (2x) cos x2
obtain that y 0 = − sin x (2x) sinsin
= −2x(sinsinx2 x+cos
= sin−2x
2 x2
2 x2 .
2
(b) The product rule with f (x) = x2 and g(x) = cos x2 and the chain for g 0 (x) = − sin x2 (2x)
so that y 0 = 2x cos x2 − sin x2 (2x)(x2 ) = 2x cos x2 − 2x3 sin x2 .
(c) Product and chain: y 0 = 3 cos(3x + 2) cos(2x − 3) − 2 sin(2x − 3) sin(3x + 2).
(d) Use the chain rule with outer f = sin−1 (u) and inner g = 2x. Derivative of the outer with
2
the inner unchanged is √ 1 2 and the derivative of the inner is 2. Thus y 0 = √1−4x
2.
1−(2x)
0
(e) Product and chain: y = a cos(ax) cos(bx) − b sin(bx) sin(ax).
(f) Use the chain rule with outer f = sin−1 (u) and inner g = x4 . Derivative of the outer with
3
.
the inner unchanged is √ 1 4 2 and the derivative of the inner is 4x3 . Thus y 0 = √4x
1−x8
1−(x )
(g) Use the product rule. y 0 = 2x cos−1 x −
2
√x
.
1−x2
(h) Use the chain rule with outer f = tan−1 (u) and inner g = ex . Derivative of the outer with
ex
the inner unchanged is 1+(e1 x )2 and the derivative of the inner is ex . Thus y 0 = 1+e
2x .
(i) Use the chain rule with outer f = eu and inner g = tan−1 x. Derivative of the outer
−1
1
with the inner unchanged is eu = etan x and the derivative of the inner is 1+x
2 . Thus
y0 =
−1
etan x
.
1+x2
3. (a) Use the substitution u = 3x + 1. The antiderivative is
(b)
(c)
(d)
(e)
1
3
sin(3x + 1) + c.
Use the substitution u = x . The antiderivative is −1
cos(x2 ) + c.
2
R
⇒ du = π5 dx ⇒ 5du
= dx. The integral becomes (9 +
Use the substitution
u = πx
5
π
R
2 sin u) 5du
= π5 (9 + 2 sin u)du = π5 (9u − 2 cos u) + c = π5 (9 πx
− 2 cos πx
) + c = 9x −
π
5
5
10
πx
cos
+
c.
π
5
The antiderivative is sin−1 x. Substituting the bounds, you obtain sin−1 (1) − sin−1 (0) = π2 .
−1
1
Note that the integral function has the form √1−u
u with 4x2 being
2 that yields sin
2
2
2
2
2
2
2
u . This tells you that 4x = u (careful: not u but u ). With
R 1u =du4x ,1 you−1can have
du
u = 2x. Thus, du = 2dx ⇒ 2 = dx. Substitute and obtain √1−u2 2 = 2 sin u + c =
1
sin−1 (2x) + c.
2
1
−1
x with 4x2 being u2 .
(f) Note that the integral function has the form 1+u
2 that yields tan
This tells you that 4x2R = u2 . So, you can take u = 2x. Thus, du = 2dx ⇒ du
= dx.
2
1
1
−1
−1
Substitute and obtain u21+1 du
=
tan
u
+
c
=
tan
(2x)
+
c.
2
2
2
R
R
1
1
(g) To obtain the form 1+u2 , factor 4 out of the denominator first. x21+4 dx =
dx =
2
4( x4 +1)
R
2
1
1
1
2
dx. Note that the integral function is of the form 1+u
= x4 . So, you
2 , with u
x2
4
+1
4
R
can take u = x2 . Thus, du = dx
⇒ 2du = dx. Substitute and obtain 14 u21+1 2du =
2
1
tan−1 u + c = 12 tan−1 ( x2 ) + c.
2
x
1
(h) Note that the function xx+1
The first
2 +4 is the sum x2 +4 + x2 +4 . Integrate both terms.
2
= dx.
integralR can be evaluated
the substitution u = x + 4 ⇒ du = 2xdx ⇒ du
2x
R x du using
x
1
Thus, x2 +4 dx = u 2x = 2 ln |u| = 21 ln(x2 + 4). The second integral reduces to the
4
previous problem with substitution u =
is 21 ln(x2 + 4) + 12 tan−1 ( x2 ) + c.
x
2
and solution
1
2
tan−1 ( x2 ). Thus, the final answer
x
3
(i) Note that the function xx+3
The first
2 +9 is the sum x2 +9 + x2 +9 . Integrate both terms.
2
integralR can be evaluated
the substitution u = x + 9 ⇒ du = 2xdx ⇒ du
= dx.
2x
R x du using
1
x
Thus, x2 +9 dx = u 2x = 2 ln |u| = 12 ln(x2 + 9).
1
The second integral is similar to problem 7. To obtain the form 1+u
2 , factor 9 out of the
R 3
denominator first. You can also factor 3 out so that it is out of your way.
dx =
x2 +9
R 1
x2
3
1
2
dx. Note that the integral function is of the form 1+u2 , with u = 9 . So, you
x2
9
+1
9
R
can take u = x3 . Thus, du = dx
⇒ 3du = dx. Substitute and obtain 13 u21+1 3du =
3
tan−1 u + c = tan−1 ( x3 ) + c. Thus, the final answer is 12 ln(x2 + 9) + tan−1 ( x3 ) + c.
(j) To apply the ideas for solving previous problems to this one, you want to complete the
denominator to squares first (that is: to write the quadratic of the form ax2 + bx + c in
the form (px + q)2 + r2 .). The denominator x2 + 4x + 13 is equal to x2 + 2(2)x + 22 + 9.
Note that the first three terms are equal to (x + 2)2 . Thus, the denominator is equal to
(x + 2)2 + 9.
R x+6
dx as the sum of two
This tells you that you want to evaluate the integral x2 +4x+13
2
integrals. For the first you will take the whole denominator (x+2) +9 for u. This indicates
du
how to decompose the numerator so that du = 2(x + 2)dx ⇒ 2(x+2)
= dx cancels the xR x+6
R x+2+4
R x+2
R
terms in the numerator. Thus x2 +4x+13 dx = (x+2)2 +9 dx = (x+2)2 +9 dx+ (x+2)4 2 +9 dx.
R
For the first integral you obtain u1 du
= 12 ln u = 12 ln((x + 2)2 + 9).
2
Reduce the second integral to formula u21+1 that will yield tan−1 (u). Note that the denominator (x + 2)2 + 9 is equal to 9[( x+2
)2 + 1]. So, for the second integral, you
use the
3
R can
1
dx
4
⇒
du
=
⇒
3du
=
dx.
This
integral
becomes
=
substitution u = x+2
dx =
3
3
9
( x+2
)2 +1
3
R
4
3 u21+1 du = 43 tan−1 u = 43 tan−1 x+2
. Thus, the final answer is 21 ln((x + 2)2 + 9) +
9
3
4
tan−1 x+2
+ c.
3
3
4. Find intersections. The equation sin x = cos x
has solutions x = π4 and x = 5π
by prob4
lem 1 (f). Graph the functions and note
that the area consists of 3 regions as in the
graph on the right. A = A1 + A2 + A3 =
R π/4
R 5π/4
(cos
x
−
sin
x)dx
+
(sin x − cos x)dx +
π/4
R02π
π/4
(cos x − sin x)dx = (sin x + cos x)|0 +
5π/4
5π/4
(− cos x − sin x)|π/4 +(sin x + cos x)|2π
5π/4 =
√
4 2 ≈ 5.657.
5