Question

∫

x
x
2
−100
​

​
dx,x>0 a) First, select one of the following trig identities to be the backbone of your substitution: 1−sin
2
(θ)=cos
2
(θ)sec
2
(θ)−1=tan
2
(θ)tan
2
(θ)+1=sec
2
(θ) Hint: Choose the trigonometric identity whose lett side most resembles x
2
−100. The right side of each identity is a perfect square which will ultimately help? you to eliminate the radical and simplity the integrall b) Next, factor out 100 from the radical to obtain an expression of the form
x
2
−100
​
=10
A
2
−1
​
. What is A? A= [a] 족. FORMATTING: The expression for A will depend on x. Since we cannor type the Grook lotter θ (thota? in Moblus answers, we will use the variablo Q in place of the traditional θ, for the remaining parts of this question. c) Choose a substitution for x so that the expression A
2
−1 (in terms of x ) is exaclly the lett side of the trig identity from a).
x=10sec
2
(Q)
x=10sin(Q)
x=
10
1
​
sin(Q)
x=
10
1
​
sec(Q)
x=10sin
2
(Q)
x=10sec(Q)
x=
10
1
​
tan
2
(Q)
x=10tan(Q)
x=10tan2(Q)
​

( where 0≤Q<
2
π
​
or
2
π
​
( where −
2
π
​
≤Q≤
2
π
​
)
( where −
2
π
​
≤Q≤
2
π
​
)
( where 0≤Q<
2
π
​
or
2
π
​
( where −
2
π
​
≤Q≤
2
π
​
)
( where 0≤Q<
2
π
​
or
2
π
​
(where −
2
π
​
2
π
​
)
(where −
2
π
​
2
π
​
)
(where −
2
π
​
2
π
​
)
​
d) Using the substitution from c), write Q in terms of x using the appropriate inverse trig function. FORMATTING: Pay attention to Moblus symitax for inverse trig functions/ 0.9 sin
−1
(a) is witten as arcsin(a), sec
−1
(a) is writiten as arcsoc(a), and tan
−1
(a) is witten as arctan(a), olc. Q= e) Using the substitution from c), write each trig ratio in terms of x. Hint: You might find it holpful to draw a right triangle with acute anglo Q and sides based on your subsitution from o). FORMATTING: Write
a
​
as sqrt(a) and use calculator notation. Your answors should not involve trig funtions nor the varliable Q. sin(Q)= cos(Q)= tan(Q)= csc(Q)= sec(Q)= cot(Q)= f) To complete this trig substitution, find
dQ
dx
​
= (4) so that you can substitute dx from the original intogral with dx= (3) dQ. FOAMATTING: Do not include dQ in your answer; it is already written for you. g) Next, use the work youve done and apply the substitution for x and dx. Then, use the right side of your cleverly-chosen trig identity to elliminate the square root and simplity the integrand. This should translorm the original integral, witten in terms of x, into a new integral in terms of Q which does not ∫
x
x
2
−100
​

​
dx=∫g(Q)dQ The now integrand is g(Q)= [4. FORMATTING: Your answer for g(Q) should not include any radicals (i.e. do not use sqrto). It should not include absolute value (ie. do not use abso). it may be a mathematical exprossion involving trig functions and the variable Q. Pay attontion to Moblus syntax for powers of trig functions, og sin
2
(a) must bo writtan as (sin(a))
∧
2 h) Now, integrate with respect to Q to obtain ∫g(Q) dQ = 선견 +C Hint: Using a ting identify may hep you evatuate this integrat. strong) FORMATTING: Your answer should depend on the variable Q. Do not include the constant of intogration in your answer, "t C
∗
is alroady written for you. Hint: You might find it helpful to draw a right trianglo with acute anglo Q and sides based on your subsitution from o) FORMATTING: Writo
a
​
as sqrt(a) and uso calculator notation. Your answers should not involvo trig funtions nor the variablo Q. sin(Q)= cos(Q)= tan(Q)= f) To complete this trig substitution, find
dQ
dx
​
= (19 20 so that you can substitute dx from the original integral with dx= (3) LQ. FORMATTING: Do not include dQ in your answer; it is atroady witten for you. 9) Next, use the work youve done and apply the substitution for x and dx. Then, use the right side of your cleverty-chosen trig identity to eliminate the square root and simplity the integrand. This should transtorm the orginal integral, written in terms of x, into a new integral in terms of Q which does not contain any square rootst ∫
x
x
2
−100
​

​
dx=∫g(Q)dQ The new integrand is g(Q)= (4) FOAMATTING: Your answer for g(Q) should not include any radicals (l.e. do not use sqrt). It should not include absolute value (ie do not use abso). It may be a mathematical expression involing trig functions and the variable Q. Pay attention fo Moblus syntax for powers of trig functions, eg sin2 ( a ) must be written as (sin(a))
∧
2 h) Now, integrate with respect to Q to obtain ∫g(Q)dQ= (4) 녁 +C. Hint: Using a trig identily may help you evaluate this integral. strong. FORMATTING: Your answer should depend on the variable Q. Do not inctude the constant of integration in your answer: "+ C " is alroady written for you 1) Finally, rewnile your answer from h in terms of the original variable x : Hint Parts of and or e ) may be useful. Final answer: ∫
x
x
2
−100
​

​
dx= 현 +C. FORMATTING: Do not include the constant of integration in your answor: "t
−1
is alleady witten for you.

Answer 1

In this section we use trigonometric identities to integrate certain combinations of trigonometric functions. We start with powers of sine and cosine.

EXAMPLE 1 Evaluate .

SOLUTION Simply substituting isn’t helpful, since then . In order

to integrate powers of cosine, we would need an extra factor. Similarly, a power of

sine would require an extra factor. Thus, here we can separate one cosine factor

and convert the remaining factor to an expression involving sine using the identity

:

We can then evaluate the integral by substituting , so and

In general, we try to write an integrand involving powers of sine and cosine in a form

where we have only one sine factor (and the remainder of the expression in terms of

cosine) or only one cosine factor (and the remainder of the expression in terms of sine).

The identity enables us to convert back and forth between even powers

of sine and cosine.

EXAMPLE 2 Find

SOLUTION We could convert to , but we would be left with an expression in

terms of with no extra factor. Instead, we separate a single sine factor and

rewrite the remaining factor in terms of :

Substituting , we have and so

1

3 cos3

x 2

5 cos5

x 1

7 cos7

x C

u3

3 2

u5

5

u7

7 C

y 1 u2

2

u2 du y u2 2u4 u6

du

y 1 cos2

x

2 cos2

x sin x dx

y sin5

x cos2

x dx y sin2

x

2 cos2

x sin x dx

u cos x du sin x dx

sin5

x cos2

x sin2

x

2

cos2

x sin x 1 cos2

x

2

cos2

x sin x

sin cos x 4

x

sin x cos x

1 sin2 cos x 2

x

y sin5

x cos2

x dx

sin2

x cos2

x 1

sin x 1

3 sin3

x C

y 1 u2 du u 1

3 u3 C

y cos3

x dx y cos2

x cos x dx y 1 sin2

x cos x d

Explanation: