integrals by trigonometric substitution - intuitive calculus · integrals by trigonometric...
TRANSCRIPT
Integrals by Trigonometric Substitution
Let’s say we want to find this integral:
1. The denominator looks like a trig identity:
Integrals by Trigonometric Substitution
Let’s say we want to find this integral:∫dx√
1− x2
1. The denominator looks like a trig identity:
Integrals by Trigonometric Substitution
Let’s say we want to find this integral:∫dx√
1− x2
1. The denominator looks like a trig identity:
Integrals by Trigonometric Substitution
Let’s say we want to find this integral:∫dx√
1− x2
1. The denominator looks like a trig identity:
cos2 t = 1− sin2 t
Integrals by Trigonometric Substitution
Let’s say we want to find this integral:∫dx√
1− x2
1. The denominator looks like a trig identity:
cos2 t = 1− sin2 t
We make the substitution:
Integrals by Trigonometric Substitution
Let’s say we want to find this integral:∫dx√
1− x2
1. The denominator looks like a trig identity:
cos2 t = 1− sin2 t
We make the substitution:
x2 = sin2 t
Integrals by Trigonometric Substitution
Let’s say we want to find this integral:∫dx√
1− x2
1. The denominator looks like a trig identity:
cos2 t = 1− sin2 t
We make the substitution:
x2 = sin2 t ⇒ x = sin t
Integrals by Trigonometric Substitution
Let’s say we want to find this integral:∫dx√
1− x2
1. The denominator looks like a trig identity:
cos2 t = 1− sin2 t
We make the substitution:
x2 = sin2 t ⇒ x = sin t
We need to find dx :
Integrals by Trigonometric Substitution
Let’s say we want to find this integral:∫dx√
1− x2
1. The denominator looks like a trig identity:
cos2 t = 1− sin2 t
We make the substitution:
x2 = sin2 t ⇒ x = sin t
We need to find dx :
dx
dt=
d
dt(sin t)
Integrals by Trigonometric Substitution
Let’s say we want to find this integral:∫dx√
1− x2
1. The denominator looks like a trig identity:
cos2 t = 1− sin2 t
We make the substitution:
x2 = sin2 t ⇒ x = sin t
We need to find dx :
dx
dt=
d
dt(sin t) = cos t
Integrals by Trigonometric Substitution
Let’s say we want to find this integral:∫dx√
1− x2
1. The denominator looks like a trig identity:
cos2 t = 1− sin2 t
We make the substitution:
x2 = sin2 t ⇒ x = sin t
We need to find dx :
dx
dt=
d
dt(sin t) = cos t
dx = cos tdt
Integrals by Trigonometric Substitution
So, we can make the substitution:
Integrals by Trigonometric Substitution
So, we can make the substitution:
x = sin t dx = cos tdt
Integrals by Trigonometric Substitution
So, we can make the substitution:
x = sin t dx = cos tdt
∫dx√
1− x2
Integrals by Trigonometric Substitution
So, we can make the substitution:
x = sin t dx = cos tdt
∫dx√
1− x2dx =
∫cos tdt√1− sin2 t
Integrals by Trigonometric Substitution
So, we can make the substitution:
x = sin t dx = cos tdt
∫dx√
1− x2dx =
∫cos tdt
������:
cos t√1− sin2 t
Integrals by Trigonometric Substitution
So, we can make the substitution:
x = sin t dx = cos tdt
∫dx√
1− x2dx =
∫cos tdt
������:
cos t√1− sin2 t
=
∫cos t
cos tdt
Integrals by Trigonometric Substitution
So, we can make the substitution:
x = sin t dx = cos tdt
∫dx√
1− x2dx =
∫cos tdt
������:
cos t√1− sin2 t
=
∫���cos t
���cos tdt =
Integrals by Trigonometric Substitution
So, we can make the substitution:
x = sin t dx = cos tdt
∫dx√
1− x2dx =
∫cos tdt
������:
cos t√1− sin2 t
=
∫���cos t
���cos tdt =
∫dt
Integrals by Trigonometric Substitution
So, we can make the substitution:
x = sin t dx = cos tdt
∫dx√
1− x2dx =
∫cos tdt
������:
cos t√1− sin2 t
=
∫���cos t
���cos tdt =
∫dt = t + C
Integrals by Trigonometric Substitution
So, we can make the substitution:
x = sin t dx = cos tdt
∫dx√
1− x2dx =
∫cos tdt
������:
cos t√1− sin2 t
=
∫���cos t
���cos tdt =
∫dt = t + C
Now we need to substitute back:
Integrals by Trigonometric Substitution
So, we can make the substitution:
x = sin t dx = cos tdt
∫dx√
1− x2dx =
∫cos tdt
������:
cos t√1− sin2 t
=
∫���cos t
���cos tdt =
∫dt = t + C
Now we need to substitute back:
x = sin t
Integrals by Trigonometric Substitution
So, we can make the substitution:
x = sin t dx = cos tdt
∫dx√
1− x2dx =
∫cos tdt
������:
cos t√1− sin2 t
=
∫���cos t
���cos tdt =
∫dt = t + C
Now we need to substitute back:
x = sin t ⇒ t = arcsin x
Integrals by Trigonometric Substitution
So, we can make the substitution:
x = sin t dx = cos tdt
∫dx√
1− x2dx =
∫cos tdt
������:
cos t√1− sin2 t
=
∫���cos t
���cos tdt =
∫dt = t + C
Now we need to substitute back:
x = sin t ⇒ t = arcsin x
So:
Integrals by Trigonometric Substitution
So, we can make the substitution:
x = sin t dx = cos tdt
∫dx√
1− x2dx =
∫cos tdt
������:
cos t√1− sin2 t
=
∫���cos t
���cos tdt =
∫dt = t + C
Now we need to substitute back:
x = sin t ⇒ t = arcsin x
So: ∫dx√
1− x2dx = arcsin x + C
Integrals by Trigonometric Substitution
So, we can make the substitution:
x = sin t dx = cos tdt
∫dx√
1− x2dx =
∫cos tdt
������:
cos t√1− sin2 t
=
∫���cos t
���cos tdt =
∫dt = t + C
Now we need to substitute back:
x = sin t ⇒ t = arcsin x
So: ∫dx√
1− x2dx = arcsin x + C
Why Trigonometric Substitution Works
I Trigonometric substitution allows us to simplify radicals,because of the identities:
I When you have an expression of the form:
I a2 − x2 ⇒ Use x = a sin t
I a2 + x2 ⇒ Use x = a tan t
I x2 − a2 ⇒ Use x = a sec t
Why Trigonometric Substitution Works
I Trigonometric substitution allows us to simplify radicals,because of the identities:
I When you have an expression of the form:
I a2 − x2 ⇒ Use x = a sin t
I a2 + x2 ⇒ Use x = a tan t
I x2 − a2 ⇒ Use x = a sec t
Why Trigonometric Substitution Works
I Trigonometric substitution allows us to simplify radicals,because of the identities:
1− cos2 t = sin2 t
I When you have an expression of the form:
I a2 − x2 ⇒ Use x = a sin t
I a2 + x2 ⇒ Use x = a tan t
I x2 − a2 ⇒ Use x = a sec t
Why Trigonometric Substitution Works
I Trigonometric substitution allows us to simplify radicals,because of the identities:
1− cos2 t = sin2 t
1 + tan2 t = sec2 t
I When you have an expression of the form:
I a2 − x2 ⇒ Use x = a sin t
I a2 + x2 ⇒ Use x = a tan t
I x2 − a2 ⇒ Use x = a sec t
Why Trigonometric Substitution Works
I Trigonometric substitution allows us to simplify radicals,because of the identities:
1− cos2 t = sin2 t
1 + tan2 t = sec2 t
I When you have an expression of the form:
I a2 − x2 ⇒ Use x = a sin t
I a2 + x2 ⇒ Use x = a tan t
I x2 − a2 ⇒ Use x = a sec t
Why Trigonometric Substitution Works
I Trigonometric substitution allows us to simplify radicals,because of the identities:
1− cos2 t = sin2 t
1 + tan2 t = sec2 t
I When you have an expression of the form:I a2 − x2 ⇒ Use x = a sin t
I a2 + x2 ⇒ Use x = a tan t
I x2 − a2 ⇒ Use x = a sec t
Why Trigonometric Substitution Works
I Trigonometric substitution allows us to simplify radicals,because of the identities:
1− cos2 t = sin2 t
1 + tan2 t = sec2 t
I When you have an expression of the form:I a2 − x2 ⇒ Use x = a sin t And you’ll get:
a2 − x2 = a2(1− sin2 t) = a2 cos2 t
I a2 + x2 ⇒ Use x = a tan t
I x2 − a2 ⇒ Use x = a sec t
Why Trigonometric Substitution Works
I Trigonometric substitution allows us to simplify radicals,because of the identities:
1− cos2 t = sin2 t
1 + tan2 t = sec2 t
I When you have an expression of the form:I a2 − x2 ⇒ Use x = a sin t And you’ll get:
a2 − x2 = a2(1− sin2 t) = a2 cos2 t
I a2 + x2 ⇒ Use x = a tan t
I x2 − a2 ⇒ Use x = a sec t
Why Trigonometric Substitution Works
I Trigonometric substitution allows us to simplify radicals,because of the identities:
1− cos2 t = sin2 t
1 + tan2 t = sec2 t
I When you have an expression of the form:I a2 − x2 ⇒ Use x = a sin t And you’ll get:
a2 − x2 = a2(1− sin2 t) = a2 cos2 t
I a2 + x2 ⇒ Use x = a tan t And you’ll get:a2 + x2 = a2(1 + tan2 t) = a2 sec2 t
I x2 − a2 ⇒ Use x = a sec t
Why Trigonometric Substitution Works
I Trigonometric substitution allows us to simplify radicals,because of the identities:
1− cos2 t = sin2 t
1 + tan2 t = sec2 t
I When you have an expression of the form:I a2 − x2 ⇒ Use x = a sin t And you’ll get:
a2 − x2 = a2(1− sin2 t) = a2 cos2 t
I a2 + x2 ⇒ Use x = a tan t And you’ll get:a2 + x2 = a2(1 + tan2 t) = a2 sec2 t
I x2 − a2 ⇒ Use x = a sec t
Why Trigonometric Substitution Works
I Trigonometric substitution allows us to simplify radicals,because of the identities:
1− cos2 t = sin2 t
1 + tan2 t = sec2 t
I When you have an expression of the form:I a2 − x2 ⇒ Use x = a sin t And you’ll get:
a2 − x2 = a2(1− sin2 t) = a2 cos2 t
I a2 + x2 ⇒ Use x = a tan t And you’ll get:a2 + x2 = a2(1 + tan2 t) = a2 sec2 t
I x2 − a2 ⇒ Use x = a sec t And you’ll get:x2 − a2 = a2(sec2 t − 1) = a2 tan2 t
