SHANKAR VENKATARAMANI
1. Trigonometric substitutions
Integrals involving the square roots of quadratic expressions can often be evaluated by a judicious substitution using trigonometric functions. I will illustrate this method using a few examples. At the end, there are a few problems for you to practice.
Example 1. ais a constant. Find Z
1
√
a2−x2dx
Solution: The idea is to make a substitution which entirely removes the radical. The “obvious” formulaa2 = (√a2−x2)2+x2 is the Pythagorean theorem for the right triangle whose sides are√a2−x2 (related to the integrand),aandx(simpler expressions than the integrand). This relationship can be represented by the following figure, where we have introduced an angleθ.
√ a2−x2
x a
θ
Figure 1. The triangle which illustrates the substitution x = asinθ for
integrands which involve the quantity√a2−x2
The substitution that simplifies the problem is to replace all the terms which depend on
x by terms which depend on θ. In particular, from the triangle, we get
x=asin(θ) =⇒ dx=acos(θ)dθ.
Also, we have
cos(θ) =
√
a2−x2
a =⇒ p
a2−x2=acos(θ).
Consequently, the substitution x=asin(θ) in the integral yields
Z 1
√
a2−x2dx=
Z 1
acos(θ)acos(θ)dθ
=
Z dθ
=θ+C.
Note that, the substitution “worked” in that it removed the radical√in the first line after the substitution. At the end, we have the integral in terms of θ and we now have to back substitute for θ in terms of x. From the definition x = asin(θ) (or from the triangle in Fig. 1) we see that
sin(θ) = x
a =⇒ θ= arcsin x
a
.
Back substitution now yields
Z 1
√
a2−x2dx=θ+C= arcsin
x a
+C.
Example 2. Find
Z √4−x2
x dx
Solution: The integrand involves the radical√4−x2 =√22−x2. This suggests consid-ering the substitution x = 2 sinθ or equivalently, right triangle with sides √4−x2 and x, which has a hypotenuse 2.
√
4−x2
x
2
θ
Figure 2. The triangle which illustrates the substitution x= 2 sinθ.
From the substitution (or from the triangle), we get
x= 2 sin(θ) =⇒ dx= 2 cos(θ)dθ.
Also, we have
cos(θ) =
√
4−x2
2 =⇒
p
Consequently, the substitution x= 2 sin(θ) in the integral yields
Z √4−x2
x dx=
Z 2 cos(θ)
2 sin(θ)2 cos(θ)dθ
= 2
Z cos2(θ)
sin(θ) dθ
= 2
Z 1−sin2(θ)
sin(θ) dθ
= 2
Z 1
sin(θ)dθ−2
Z
sin(θ)dθ
= ln
1−cos(θ) 1 + cos(θ)
+ 2 cos(θ) +C.
where the final expression is obtained from the table of integrals. From the triangle cos(√ θ) = 4−x2
2 , so back substitution now yields
Z √
4−x2
x dx= ln
1−cos(θ) 1 + cos(θ)
+ 2 cos(θ) +C
= ln
2−√4−x2
2 +√4−x2
+p4−x2+C.
Example 3 (Completing squares). Find Z
(8−2x−x2)−3/2dx
Solution: If you see an expression that involves a square root and a quadratic, do not factor the quadratic. Instead, complete squares. Also, before completing squares, factor out the coefficient in front of the x2. In this example, we have
8−2x−x2=−
x2+ 2x−8
factoring −1, the coefficient of the x2
=−
x2+ 2x+ 1−1−8
=−
(x+ 1)2−9
= 32−(x+ 1)2.
Consequently,
Z
(8−2x−x2)−3/2dx=
Z
1
(p32−(x+ 1)2)3dx.
This suggests the substitution,x+ 1 =uand u= 3 sin(θ). Then, du=dxand from the triangle we get
u= 3 sin(θ) =⇒ dx= 3 cos(θ)dθ, cos(θ) =
√
9−u2
2 =⇒
p
√
8−2x−x2 =√9−u2
x+ 1 =u
3
θ
Figure 3. The triangle which illustrates the substitutionx+ 1 =u= 3 sinθ.
Substituting for x by u, and then foru by θ, we get
Z
(8−2x−x2)−3/2dx=
Z
du
(√9−u2)3
=
Z
3 cos(θ)dθ
(3 cos(θ))3 dθ
= 1 9
Z
1 cos2(θ)dθ
= 1
9tan(θ) +C. From the triangle
tan(θ) = √ u
9−u2 =
x+ 1
√
8−2x−x2, so back substitution now yields
Z
(8−2x−x2)−3/2dx= x+ 1
9√8−2x−x2 +C.
Example 4. ais a constant. Find
Z 1
√
a2+x2dx
Solution: Again, the idea is to make a substitution which entirely removes the radical. The “obvious” formula (√a2+x2)2 =a2+x2 is the Pythagorean theorem for the right triangle whose sides areaandxand whose hypotenuse is√a2+x2(related to the integrand),aand
x. This relationship can be represented by the following figure, where we have introduced an angle θ.
a
x √
a2+x2
θ
Figure 4. The triangle which illustrates the substitution x=atanθ.
The substitution that simplifies the problem is to replace all the terms which depend on
x by terms which depend on θ. In particular, from the triangle, we get
x=atan(θ) =⇒ dx= a cos2(θ)dθ.
Also, we have
cos(θ) = √ a
a2+x2 =⇒
p
a2+x2= a cos(θ).
Consequently, the substitution x=atan(θ) in the integral yields
Z
1
√
a2+x2dx=
Z
cos(θ)
a
adθ
cos2(θ)
=
Z dθ
cos(θ)
= 1 2ln
1 + sin(θ) 1−sin(θ)
+C.
where the final expression is obtained from the table of integrals.
to back substitute for θin terms ofx. From the definition x=atan(θ) and the triangle in Fig. 4) we see that
sin(θ) = √ x a2+x2.
Back substitution now yields
Z
dx √
a2+x2 = 1 2ln
√
a2+x2+x
√
a2+x2−x
+C.
Although it looks different, this expression is equal to the formula in the integral table.
Example 5. Find
Z dx
x(16 +x2)
Solution: The integrand is a rational function so this problem can be solved by the method of partial fractions. We can also do it by trigonometric substitution by recognizing that the integrand has a facto 16 +x2= 42+x2 which suggests the substitutionx= 4 tanθ.
4
x √
16 +x2
θ
Figure 5. The triangle which illustrates the substitutionx= 4 tanθ.
From the substitution and the corresponding triangle we get
x= 4 tan(θ) =⇒ dx= 4 cos2(θ)dθ.
Also, we have
cos(θ) = √ 4
16 +x2 =⇒
p
Consequently, the substitution x= 4 tan(θ) in the integral yields
Z 1
x(16 +x2)dx=
Z 1
4 tan(θ)
cos2(θ) 16
4dθ
cos2(θ)
= 1 16
Z dθ
tan(θ)
= 1 16
Z cos(θ)dθ
sin(θ)
= 1
16ln|sin(θ)|+C.
where the final expression is obtained by th substitution u= sin(θ) =⇒ du = cos(θ)dθ. Back substituting using the triangle we get
Z
1
x(16 +x2)dx= 1 16ln
x √
16 +x2
+C = 1 32
ln|x2| −ln|16 +x2|
+C.
2. Problems
You can check your answers with Wolfram Alpha! (1) ais a constant. Find the antiderivatives (2 cases!)
Z 1
x√a2±x2 dx. (2) Find
Z
(4x2+ 1)−3/2dx.
(3) Find
Z p
x2+ 8x+ 25dx.
(4) Find
Z p
3 + 4x−4x2dx
(5) Find
Z 1
√
e2x+ 1dx.
