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Trigonometric Substitution Integrals involving a^2 − x^2 Integrals involving x^2 + a^2 Integrals involving x^2 − a^2
Trigonometric Substitution
To solve integrals containing the following expressions;
p a^2 − x^2
p x^2 + a^2
p x^2 − a^2 ,
it is sometimes useful to make the following substitutions:
q^ Expression^ Substitution^ Identity a^2 − x^2 x = a sin θ, − π 2 ≤ θ ≤ π 2 or θ = sin−^1 xa 1 − sin^2 θ = cos^2 θ p q^ a^2 +^ x^2 x^ =^ a^ tan^ θ,^ −^ π^2 ≤^ θ^ ≤^ π^2 or^ θ^ = tan−^1 xa^ 1 + tan^2 θ^ = sec^2 θ x^2 − a^2 x = a sec θ, 0 ≤ θ < π 2 or π ≤ θ < 32 π or θ = sec−^1 xa sec^2 θ − 1 = tan^2 θ
Note The calculations here are much easier if you use the substitution in
reverse: x = a sin θ as opposed to θ = sin
− 1 x a.
Trigonometric Substitution Integrals involving a^2 − x^2 Integrals involving x^2 + a^2 Integrals involving x^2 − a^2
Integrals involving
a
− x
We make the substitution x = a sin θ, −
π 2 ≤^ θ^ ≤^
π √^2 ,^ dx^ =^ a^ cos^ θdθ, a^2 − x^2 =
p a^2 − a^2 sin^2 θ = a| cos θ| = a cos θ (since − π 2 ≤ θ ≤ π 2 by choice. )
Example Z x
3 √ 4 − x^2
dx
Trigonometric Substitution Integrals involving a^2 − x^2 Integrals involving x^2 + a^2 Integrals involving x^2 − a^2
Integrals involving
a
− x
We make the substitution x = a sin θ, −
π 2 ≤^ θ^ ≤^
π √^2 ,^ dx^ =^ a^ cos^ θdθ, a^2 − x^2 =
p a^2 − a^2 sin^2 θ = a| cos θ| = a cos θ (since − π 2 ≤ θ ≤ π 2 by choice. )
Example Z x
3 √ 4 − x^2
dx
I (^) Let x = 2 sin θ, dx = 2 cos θdθ,
4 − x^2 =
p 4 − 4 sin
2 θ = 2 cos θ.
I
R
√x^3 dx 4 −x^2
R (^) 8 sin (^3) θ(2 cos θdθ)
2 cos θ =^
R
8 sin
3 θdθ =
R
8 sin
2 θ sin θ dθ =
R
(1 − cos
2 θ) sin θ dθ.
Trigonometric Substitution Integrals involving a^2 − x^2 Integrals involving x^2 + a^2 Integrals involving x^2 − a^2
Integrals involving
a
− x
We make the substitution x = a sin θ, −
π 2 ≤^ θ^ ≤^
π √^2 ,^ dx^ =^ a^ cos^ θdθ, a^2 − x^2 =
p a^2 − a^2 sin^2 θ = a| cos θ| = a cos θ (since − π 2 ≤ θ ≤ π 2 by choice. )
Example Z x
3 √ 4 − x^2
dx
I (^) Let x = 2 sin θ, dx = 2 cos θdθ,
4 − x^2 =
p 4 − 4 sin
2 θ = 2 cos θ.
I
R
√x^3 dx 4 −x^2
R (^) 8 sin (^3) θ(2 cos θdθ)
2 cos θ =^
R
8 sin
3 θdθ =
R
8 sin
2 θ sin θ dθ =
R
(1 − cos
2 θ) sin θ dθ.
I (^) Let w = cos θ, dw = − sin θ dθ,
Z
(1−cos
2 θ) sin θ dθ = − 8
Z
(1−w
2 ) dw = 8
Z
(w
2 −1) dw =
8 w
3
− 8 w +C
8(cos θ)^3 3 −^ 8 cos^ θ^ +^ C^ =^
8(cos(sin−^1 x 2 ))^3 3 −^ 8 cos(sin
− 1 x 2 ) +^ C^.
Trigonometric Substitution Integrals involving a^2 − x^2 Integrals involving x^2 + a^2 Integrals involving x^2 − a^2
Integrals involving
a
− x
Z
x^3 √ 4 − x^2
dx =
8(cos(sin
− 1 x 2 ))
3
− 8 cos(sin
− 1 x 2
) + C.
I (^) To get an expression for cos(sin−^1 x 2 ), we use an appropriate triangle
4 - x^2
θ
x
2
From the triangle, we get
cos(sin
− 1 x 2 ) =
4 −x^2 2
Trigonometric Substitution Integrals involving a^2 − x^2 Integrals involving x^2 + a^2 Integrals involving x^2 − a^2
Integrals involving
a
− x
Z
x^3 √ 4 − x^2
dx =
8(cos(sin
− 1 x 2 ))
3
− 8 cos(sin
− 1 x 2
) + C.
I (^) To get an expression for cos(sin−^1 x 2 ), we use an appropriate triangle
4 - x^2
θ
x
2
From the triangle, we get
cos(sin
− 1 x 2 ) =
4 −x^2 2
I (^) hence
R (^) x (^3) dx √ 4 −x^2
8
4 −x^2 2
3 −^8
4 −x^2 2 +^ C^ =^
(4−x^2 )^3 /^2 3 −^4
4 − x^2 + C
Trigonometric Substitution Integrals involving a^2 − x^2 Integrals involving x^2 + a^2 Integrals involving x^2 − a^2
Integrals involving
a
− x
Example
R (^) dx
x^2
9 −x^2
I (^) Let x = 3 sin θ, dx = 3 cos θdθ,
9 − x^2 =
p 9 − 9 sin^2 θ = 3 cos θ.
Trigonometric Substitution Integrals involving a^2 − x^2 Integrals involving x^2 + a^2 Integrals involving x^2 − a^2
Integrals involving
a
− x
Example
R (^) dx
x^2
9 −x^2
I (^) Let x = 3 sin θ, dx = 3 cos θdθ,
9 − x^2 =
p 9 − 9 sin^2 θ = 3 cos θ.
I
R (^) dx
x^2
9 −x^2
R (^) 3 cos θdθ (9 sin^2 θ)3 cos θ =^
R 1
9 sin^2 θ dθ^ =^
− cot θ 9 +^ C^ =^
− cot(sin−^1 x 3 ) 9 +^ C
Trigonometric Substitution Integrals involving a^2 − x^2 Integrals involving x^2 + a^2 Integrals involving x^2 − a^2
Integrals involving
a
− x
Example
R (^) dx
x^2
9 −x^2
I (^) Let x = 3 sin θ, dx = 3 cos θdθ,
9 − x^2 =
p 9 − 9 sin^2 θ = 3 cos θ.
I
R (^) dx
x^2
9 −x^2
R (^) 3 cos θdθ (9 sin^2 θ)3 cos θ =^
R 1
9 sin^2 θ dθ^ =^
− cot θ 9 +^ C^ =^
− cot(sin−^1 x 3 ) 9 +^ C
I (^) To get an expression for cot(sin−^1 x 3 ), we use an appropriate triangle
9 - x^2
θ
x
3
From the triangle, we get
cot(sin−^1 x 3 ) =
9 −x^2 x and hence
Z dx
x^2
9 − x^2
9 − x^2
9 x
+ C
I (^) Note You can also use this method to derive what you already know
Z 1 √ a^2 − x^2
dx = sin
− 1 x
a
+ C
Trigonometric Substitution Integrals involving a^2 − x^2 Integrals involving x^2 + a^2 Integrals involving x^2 − a^2
Integrals involving
x
+ a
We make the substitution x = a tan θ, −
π 2 ≤^ θ^ ≤^
π 2 ,^ dx^ =^ a^ sec
2 θdθ, √ x^2 + a^2 =
a^2 tan^2 θ + a^2 = a| sec θ| = a sec θ (since − π 2 ≤ θ ≤ π 2 by choice. )
Example (^) Z dx √ x^2 + 4
Trigonometric Substitution Integrals involving a^2 − x^2 Integrals involving x^2 + a^2 Integrals involving x^2 − a^2
Integrals involving
x
+ a
We make the substitution x = a tan θ, −
π 2 ≤^ θ^ ≤^
π 2 ,^ dx^ =^ a^ sec
2 θdθ, √ x^2 + a^2 =
a^2 tan^2 θ + a^2 = a| sec θ| = a sec θ (since − π 2 ≤ θ ≤ π 2 by choice. )
Example (^) Z dx √ x^2 + 4
I (^) Let x = 2 tan θ, − π 2 ≤^ θ^ ≤^
π 2 ,^ dx^ = 2 sec
(^2) θdθ, √ x^2 + 4 =
4 tan^2 θ + 4 = 2
sec^2 θ = 2 sec θ.
I
R (^) dx √ x^2 +
R (^) 2 sec (^2) θdθ 2 sec θ =^
R
sec θdθ = ln | sec θ + tan θ| + C =
ln | sec(tan
− (^1) x 2 ) + tan(tan
− (^1) x 2 )|^ +^ C^ = ln^ |^ sec(tan
− (^1) x 2 ) +^
x 2 |^ +^ C
Trigonometric Substitution Integrals involving a^2 − x^2 Integrals involving x^2 + a^2 Integrals involving x^2 − a^2
Integrals involving
x
+ a
We make the substitution x = a tan θ, −
π 2 ≤^ θ^ ≤^
π 2 ,^ dx^ =^ a^ sec
2 θdθ, √ x^2 + a^2 =
a^2 tan^2 θ + a^2 = a| sec θ| = a sec θ (since − π 2 ≤ θ ≤ π 2 by choice. )
Example (^) Z dx √ x^2 + 4
I (^) Let x = 2 tan θ, − π 2 ≤^ θ^ ≤^
π 2 ,^ dx^ = 2 sec
(^2) θdθ, √ x^2 + 4 =
4 tan^2 θ + 4 = 2
sec^2 θ = 2 sec θ.
I
R (^) dx √ x^2 +
R (^) 2 sec (^2) θdθ 2 sec θ =^
R
sec θdθ = ln | sec θ + tan θ| + C =
ln | sec(tan
− (^1) x 2 ) + tan(tan
− (^1) x 2 )|^ +^ C^ = ln^ |^ sec(tan
− (^1) x 2 ) +^
x 2 |^ +^ C I (^) To get an expression for sec(tan−^1 x 2 ), we use an appropriate triangle
2
θ
x
x^2 + 4
From the triangle, we get
sec(tan
− 1 x 2 ) =
x^2 + 2 and hence
Z dx √ x^2 + 4
= ln |
x^2 + 4
2
x
2
|+C.
Trigonometric Substitution Integrals involving a^2 − x^2 Integrals involving x^2 + a^2 Integrals involving x^2 − a^2
Integrals involving
x
+ a
, Completing the square.
Sometimes we can convert an integral to a form where trigonometric
substitution can be applied by completing the square.
Example Evaluate (^) Z dx √ x^2 − 4 x + 13
Trigonometric Substitution Integrals involving a^2 − x^2 Integrals involving x^2 + a^2 Integrals involving x^2 − a^2
Integrals involving
x
+ a
, Completing the square.
Sometimes we can convert an integral to a form where trigonometric
substitution can be applied by completing the square.
Example Evaluate (^) Z dx √ x^2 − 4 x + 13
I (^) x^2 − 4 x + 13 = x^2 − 2(2)x + 2^2 − 22 + 13 = (x − 2)^2 + 9.
