Calc1231W5T1 - Integrals with hyperbolic trig substitution

Friday, 24 August 2018

1:05 PM

Untitled picture.png Machine generated alternative text:
Consider the integral
1
One method of solving this integral is to use the circle trigonometric substitution
dc
Since
du
we can rewrite this integral as:
1
1
x = 4tan(u).
du.
1 + tan2 (u)
The reason this substitution works so well is because identity
cos (u) + sin (u) = 1
can be rearranged (by dividing through by cos (u)) to simplify the square root Thereby, the integral can be evaluated first as a function of u:
I = In(lsec(u) + + C
Using a right triangle with angle u and side lengths x, 4 and (4)2 + x we can evaluate tan(u) and sec(u) as a function in terms of the original variable x:
• tan(u) —
• sec(u)
So the integral can be expressed in terms of the original variable x as
1
Note: the Maple syntax for tan (u) is tan (x) A2 . The syntax for In(lxl) is In(abs(x)) , and the syntax for is sqrt (x)

Untitled picture.png Machine generated alternative text:
Consider the integral
1
But now we try the hyperbolic trigonometric substitution
dc
Since
du
we can rewrite this integral as:
1
1
= 4 sinh(u).
cosh(u)
du.
1 + sinh2(u)
The reason this substitution works so well is because the identity
cosh2 (u) — sinh2 (u) = 1
can be rearranged to simplify the square root. The integral can be evaluated first as a function of u:
1
Then as a function in terms of the original variable x:
1
Note: the Maple syntax for cosh¯l (u) is arccosh(u)
arcsinh(W4)
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Untitled picture.png Machine generated alternative text:
Consider the integral
1
But now we try the hyperbolic trigonometric substitution
dc
Since
du
we can rewrite this integral as:
1
1
= 4 sinh(u).
cosh(u)
du.
1 + sinh2(u)
The reason this substitution works so well is because the identity
cosh2 (u) — sinh2 (u) = 1
can be rearranged to simplify the square root. The integral can be evaluated first as a function of u:
1
Then as a function in terms of the original variable x:
1
Note: the Maple syntax for cosh¯l (u) is arccosh(u)
arcsinh(W4)

Untitled picture.png Machine generated alternative text:
Here is a table of circle trig substitutions and their equivalent hyperbolic trig substitutions.
Circle trig
Integral
substitution
dc
dc
dc
x — a tan(u)
a sec(u)
Hyperbolic trig
substitution
x = a tanh(u)
x — asinh(u)
x = acosh(u)
It is often simpler to use a substitution involving cos, sin or their hyperbolic equivalents cosh, Sinh rather than using a tan, sec or tanh substitution. For example; to find the integral
1
1
(4)2
We prefer to use the circle trig substitution
dc
Since
du
4cos(u)
4sin(u)
we can simplify this integral in terms of u:
1
dc.
du
Now use the identity 1
Then in terms of the original variable x:
1
1
I—sin (u)
to evaluate this integral first in terms of u:
arcsin(x/4)
Note: the Maple syntax for cos
@) is arccos (x) and the syntax for tan @) is tan (x)

Untitled picture.png Machine generated alternative text:
Here is a table of circle trig substitutions and their equivalent hyperbolic trig substitutions.
Circle trig
Integral
substitution
dc
dc
dc
x — a tan(u)
a sec(u)
Hyperbolic trig
substitution
x = a tanh(u)
x — asinh(u)
x = acosh(u)
It is often simpler to use a substitution involving cos, sin or their hyperbolic equivalents cosh, Sinh rather than using a tan, sec or tanh substitution. For example; to find the integral
1
1
(4)2
We prefer to use the circle trig substitution
dc
Since
du
4cos(u)
4sin(u)
we can simplify this integral in terms of u:
1
dc.
du
Now use the identity 1
Then in terms of the original variable x:
1
1
I—sin (u)
to evaluate this integral first in terms of u:
arcsin(x/4)
Note: the Maple syntax for cos
@) is arccos (x) and the syntax for tan @) is tan (x)

Untitled picture.png Machine generated alternative text:
Let's see what happens if we were to integrate
1
1
(4)2 _ p
using the hyperbolic trig substitution (with variable u)
Then in terms of the variable u
dc
du
dc
and the integral becomes
I
To evaluate this integral; we can to use the definition
sech(u) —
= 4 sech2 (u)
sech(u)du_
1
cosh(u)
1
2
and a second substitution
Then in terms of the variable t the integral becomes
This can be evaluated using a table of standard integrals. First in terms of t:
1
Then in terms of u:
1
Finally in terms of the original variable x:
2*arctan(t)
2*arctan(exp(u))
Note: the Maple syntax for sin
(e
1
) is arcsin (exp(x)).

Untitled picture.png Machine generated alternative text:
An alternative way to evaluate
is to use another hyperbolic trig substitution
where
With this substitution the integral becomes
1
dt
du
sech(u) du
t = sinh(u)
1
du
cosh(u)
cosn(u)
1
g(t)dt
where g(t)
1/(1+tW2)
O
Hence we can evaluate I using the table of standard integrals, first as a function oft:
1
1
arctan(t)
Then as a function of the variable u:
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Untitled picture.png Machine generated alternative text:
An alternative way to evaluate
is to use another hyperbolic trig substitution
where
With this substitution the integral becomes
1
dt
du
sech(u) du
t = sinh(u)
1
du
cosh(u)
cosn(u)
1
g(t)dt
where g(t)
1/(1+tW2)
O
Hence we can evaluate I using the table of standard integrals, first as a function oft:
1
1
arctan(t)
Then as a function of the variable u:

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