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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