=
∫ π/2
−π/2
9 cos2θdθ
=
∫ π/2
−π/2
9 2
(1+cos(2θ)) dθ
=9 2
(θ+1
2sin(2θ))
π/2
−π/2
= 9 2π.
This matches our answer from before.
We now describe in detail Trigonometric Subs tu on. This method excels when dealing with integrands that contain√
a2−x2,√
x2−a2and√ x2+a2. The following Key Idea outlines the procedure for each case, followed by more examples. Each right triangle acts as a reference to help us understand the re- la onships betweenxandθ.
Key Idea 13 Trigonometric Subs tu on (a) For integrands containing√
a2−x2: Letx=asinθ, dx=acosθdθ Thusθ=sin−1(x/a), for−π/2≤θ≤π/2.
On this interval, cosθ≥0, so
√a2−x2=acosθ
.. √
a2−x2 .
x .
a
. θ
(b) For integrands containing√ x2+a2: Letx=atanθ, dx=asec2θdθ Thusθ=tan−1(x/a), for−π/2<θ<π/2.
On this interval, secθ>0, so
√x2+a2=asecθ
.. a
.
x .
√ x2 +a2
. θ
(c) For integrands containing√ x2−a2: Letx=asecθ, dx=asecθtanθdθ
Thusθ=sec−1(x/a). Ifx/a≥1, then 0≤θ <π/2;
ifx/a≤ −1, thenπ/2<θ≤π.
We restrict our work to wherex≥a, sox/a≥1, and 0≤θ<π/2. On this interval, tanθ≥0, so
√x2−a2=atanθ
.. a
.
√x2−a2 .
x
. θ
Notes:
Example 175 Using Trigonometric Subs tu on Evaluate
∫ 1
√5+x2 dx.
S Using Key Idea 13(b), we recognizea = √
5 and setx =
√5 tanθ. This makesdx =√
5 sec2θdθ. We will use the fact that√
5+x2 =
√5+5 tan2θ=√
5 sec2θ=√
5 secθ.Subs tu ng, we have:
∫ √ 1
5+x2 dx=
∫ √ 1 5+5 tan2θ
√5 sec2θdθ
=
∫ √ 5 sec2θ
√5 secθ dθ
=
∫
secθdθ
=lnsecθ+tanθ+C.
While the integra on steps are over, we are not yet done. The original problem was stated in terms ofx, whereas our answer is given in terms ofθ. We must convert back tox.
The reference triangle given in Key Idea 13(b) helps. Withx=√
5 tanθ, we have
tanθ= x
√5 and secθ=
√x2+5
√5 . This gives
∫ 1
√5+x2 dx=lnsecθ+tanθ+C
=ln
√x2+5
√5 + x
√5 +C.
We can leave this answer as is, or we can use a logarithmic iden ty to simplify it. Note:
ln
√x√2+5 5 +√x
5
+C=ln √1
5
(√x2+5+x)+C
=ln 1
√5
+ln√x2+5+x+C
=ln√x2+5+x+C, where the ln(
1/√ 5)
term is absorbed into the constantC. (In Sec on 6.6 we will learn another way of approaching this problem.)
Notes:
Example 176 Using Trigonometric Subs tu on Evaluate∫ √
4x2−1dx.
S We start by rewri ng the integrand so that it looks like√ x2−a2 for some value ofa:
√4x2−1=
√ 4
( x2−1
4 )
=2
√ x2−
(1 2
)2
.
So we havea=1/2, and following Key Idea 13(c), we setx=12secθ, and hence dx= 12secθtanθdθ. We now rewrite the integral with these subs tu ons:
∫ √4x2−1dx=
∫ 2
√ x2−
(1 2
)2
dx
=
∫ 2
√1
4sec2θ−1 4
(1
2secθtanθ )
dθ
=
∫ √1
4(sec2θ−1)(
secθtanθ )
dθ
=
∫ √1 4tan2θ
(
secθtanθ )
dθ
=
∫ 1
2tan2θsecθdθ
= 1 2
∫ (
sec2θ−1 )
secθdθ
= 1 2
∫ (sec3θ−secθ) dθ.
We integrated sec3θin Example 172, finding its an deriva ves to be
∫
sec3θdθ= 1 2 (
secθtanθ+ln|secθ+tanθ|) +C.
∫ √Thus
4x2−1dx= 1 2
∫ (sec3θ−secθ) dθ
= 1 2
(1 2 (
secθtanθ+ln|secθ+tanθ|)
−ln|secθ+tanθ| )
+C
= 1
4(secθtanθ−ln|secθ+tanθ|) +C.
Notes:
We are not yet done. Our original integral is given in terms ofx, whereas our final answer, as given, is in terms ofθ. We need to rewrite our answer in terms ofx. Witha = 1/2, andx = 12secθ, the reference triangle in Key Idea 13(c) shows that
tanθ=√
x2−1/4 /
(1/2) =2√
x2−1/4 and secθ=2x. Thus
1 4 (
secθtanθ−lnsecθ+tanθ)+C=1 4 (
2x·2√
x2−1/4−ln2x+2√
x2−1/4)+C
=1 4 (
4x√
x2−1/4−ln2x+2√
x2−1/4)+C.
The final answer is given in the last line above, repeated here:
∫ √4x2−1dx= 1 4 (
4x√
x2−1/4−ln2x+2√
x2−1/4)+C.
Example 177 Using Trigonometric Subs tu on Evaluate
∫ √ 4−x2
x2 dx.
S We use Key Idea 13(a) witha=2,x=2 sinθ,dx=2 cosθ and hence√
4−x2=2 cosθ. This gives
∫ √ 4−x2
x2 dx=
∫ 2 cosθ
4 sin2θ(2 cosθ)dθ
=
∫
cot2θdθ
=
∫
(csc2θ−1)dθ
=−cotθ−θ+C.
We need to rewrite our answer in terms ofx. Using the reference triangle found in Key Idea 13(a), we have cotθ=√
4−x2/xandθ=sin−1(x/2). Thus
∫ √ 4−x2
x2 dx=−
√4−x2
x −sin−1 (x
2 )
+C.
Trigonometric Subs tu on can be applied in many situa ons, even those not of the form√
a2−x2,√
x2−a2or√
x2+a2. In the following example, we ap- ply it to an integral we already know how to handle.
Notes:
Example 178 Using Trigonometric Subs tu on Evaluate
∫ 1 x2+1dx.
S We know the answer already as tan−1x+C. We apply Trigono- metric Subs tu on here to show that we get the same answer without inher- ently relying on knowledge of the deriva ve of the arctangent func on.
Using Key Idea 13(b), letx=tanθ,dx=sec2θdθand note thatx2+1= tan2θ+1=sec2θ. Thus
∫ 1 x2+1dx=
∫ 1
sec2θsec2θdθ
=
∫ 1dθ
=θ+C.
Sincex=tanθ,θ=tan−1x, and we conclude that
∫ 1
x2+1dx=tan−1x+C.
The next example is similar to the previous one in that it does not involve a square–root. It shows how several techniques and iden es can be combined to obtain a solu on.
Example 179 Using Trigonometric Subs tu on Evaluate
∫ 1
(x2+6x+10)2 dx.
S We start by comple ng the square, then make the subs tu- onu=x+3, followed by the trigonometric subs tu on ofu=tanθ:
∫ 1
(x2+6x+10)2 dx=
∫ 1
((x+3)2+1)2 dx=
∫ 1
(u2+1)2 du.
Now make the subs tu onu=tanθ,du=sec2θdθ:
=
∫ 1
(tan2θ+1)2sec2θdθ
=
∫ 1
(sec2θ)2sec2θdθ
=
∫
cos2θdθ.
Notes:
Applying a power reducing formula, we have
=
∫ (1 2+1
2cos(2θ) )
dθ
= 1 2θ+1
4sin(2θ) +C. (6.2) We need to return to the variablex. Asu = tanθ,θ = tan−1u. Using the iden ty sin(2θ) = 2 sinθcosθand using the reference triangle found in Key Idea 13(b), we have
1
4sin(2θ) = 1 2
√ u
u2+1· 1
√u2+1= 1 2
u u2+1.
Finally, we return toxwith the subs tu onu=x+3. We start with the expres- sion in Equa on (6.2):
1 2θ+1
4sin(2θ) +C= 1
2tan−1u+1 2
u u2+1+C
= 1
2tan−1(x+3) + x+3
2(x2+6x+10)+C.
Sta ng our final result in one line,
∫ 1
(x2+6x+10)2 dx= 1
2tan−1(x+3) + x+3
2(x2+6x+10)+C.
Our last example returns us to definite integrals, as seen in our first example.
Given a definite integral that can be evaluated using Trigonometric Subs tu on, we could first evaluate the corresponding indefinite integral (by changing from an integral in terms ofxto one in terms ofθ, then conver ng back tox) and then evaluate using the original bounds. It is much more straigh orward, though, to change the bounds as we subs tute.
Example 180 Definite integra on and Trigonometric Subs tu on Evaluate
∫ 5 0
x2
√x2+25dx.
S Using Key Idea 13(b), we setx =5 tanθ,dx =5 sec2θdθ, and note that√
x2+25 = 5 secθ. As we subs tute, we can also change the bounds of integra on.
The lower bound of the original integral isx=0. Asx=5 tanθ, we solve for θand findθ=tan−1(x/5). Thus the new lower bound isθ=tan−1(0) =0. The
Notes:
original upper bound isx=5, thus the new upper bound isθ=tan−1(5/5) = π/4.
Thus we have
∫ 5 0
x2
√x2+25 dx=
∫ π/4 0
25 tan2θ
5 secθ 5 sec2θdθ
=25
∫ π/4 0
tan2θsecθdθ.
We encountered this indefinite integral in Example 176 where we found
∫
tan2θsecθdθ= 1 2
(secθtanθ−ln|secθ+tanθ|) .
So 25
∫ π/4 0
tan2θsecθdθ=25 2
(secθtanθ−ln|secθ+tanθ|)
π/4
0
=25 2
(√2−ln(√ 2+1))
≈6.661.
The following equali es are very useful when evalua ng integrals using Trigono- metric Subs tu on.
Key Idea 14 Useful Equali es with Trigonometric Subs tu on 1. sin(2θ) =2 sinθcosθ
2. cos(2θ) =cos2θ−sin2θ=2 cos2θ−1=1−2 sin2θ 3.
∫
sec3θdθ=1 2 (
secθtanθ+lnsecθ+tanθ)+C 4.
∫
cos2θdθ=
∫ 1 2
(1+cos(2θ)) dθ= 1
2
(θ+sinθcosθ) +C.
The next sec on introduces Par al Frac on Decomposi on, which is an alge- braic technique that turns “complicated” frac ons into sums of “simpler” frac-
ons, making integra on easier.
Notes:
Terms and Concepts
1. Trigonometric Subs tu on works on the same principles as Integra on by Subs tu on, though it can feel “ ”.
2. If one uses Trigonometric Subs tu on on an integrand con- taining√
25−x2, then one should setx= . 3. Consider the Pythagorean Iden ty sin2θ+cos2θ=1.
(a) What iden ty is obtained when both sides are di- vided by cos2θ?
(b) Use the new iden ty to simplify 9 tan2θ+9.
4. Why does Key Idea 13(a) state that√
a2−x2 = acosθ, and not|acosθ|?
Problems
In Exercises 5 – 16, apply Trigonometric Subs tu on to eval- uate the indefinite integrals.
5.
∫ √
x2+1dx
6.
∫ √
x2+4dx
7.
∫ √
1−x2dx
8.
∫ √
9−x2dx
9.
∫ √
x2−1dx
10.
∫ √
x2−16dx
11.
∫ √
4x2+1dx
12.
∫ √
1−9x2dx
13.
∫ √
16x2−1dx
14.
∫ 8
√x2+2dx
15.
∫ 3
√7−x2dx
In Exercises 17 – 26, evaluate the indefinite integrals. Some may be evaluated without Trigonometric Subs tu on.
17.
∫ √ x2−11
x dx
18.
∫ 1
(x2+1)2dx 19.
∫ x
√x2−3dx
20.
∫ x2√
1−x2dx
21.
∫ x
(x2+9)3/2 dx 22.
∫ 5x2
√x2−10dx
23.
∫ 1
(x2+4x+13)2 dx 24.
∫
x2(1−x2)−3/2dx 25.
∫ √ 5−x2
7x2 dx 26.
∫ x2
√x2+3dx
In Exercises 27 – 32, evaluate the definite integrals by mak- ing the proper trigonometric subs tu onandchanging the bounds of integra on. (Note: each of the corresponding indefinite integrals has appeared previously in this Exercise set.)
27.
∫1
−1
√1−x2dx
28.
∫8
4
√x2−16dx
29.
∫2
0
√x2+4dx
30.
∫1
−1
1 (x2+1)2dx 31.
∫1
−1
√9−x2dx