Integrals by Trigonometric Substitution - Intuitive Calculus · Integrals by Trigonometric Substitution Let’s say we want to nd this integral: Z dx p 1 x2 1.The denominator looks

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