Inverse Trig Substitution Integral

In summary, the integral \int x^{2} \sqrt{4+x^{2}} dx can be solved by using the substitution x = 2tan\theta and dx = 2sec^{2}\theta d\theta. This leads to the integral 16 \int \tan^{3}\theta \sec^{3}\theta d\theta, which can be simplified by using the replacement tan^{2}\theta = \sec^{2}\theta -1. Alternatively, if the term is x^{3} instead of x^{2}, the integral can be solved by using the substitution x = sinh(u).
  • #1
Bazzinga
45
0
[tex]\int x^{2} \sqrt{4+x^{2}} dx[/tex]

I've already subbed in:

[tex] x = 2tan\theta [/tex]
[tex] dx = 2sec^{2}\theta d\theta [/tex]

and I've gotten down to:

[tex] 16 \int tan^{3}\theta sec^{3}\theta d\theta [/tex]

But now I have noo idea what to do! Can someone give me a hint?
 
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  • #2
Hi Bazzinga! :smile:

(where did your tan3 come from? :confused:)

sorry, but you just have to slog it out, using repeated integration by parts :redface:

substituting x = sinh(u) instead would have been a lot easier :wink:
 
  • #3
I think your integral should reduce to
[tex]
16 \int tan^{2}\theta sec^{3}\theta d\theta
[/tex]

as the x2 term outside the square root will give tan2.

Replace tan2θ with sec2θ -1 and try to simplify it. I think the integrals may get a bit complicated.
 
  • #4
Bazzinga said:
...
and I've gotten down to:

[tex] 16 \int \tan^{3}\theta \sec^{3}\theta d\theta [/tex]

rock.freak667 said:
I think your integral should reduce to
[tex]
16 \int \tan^{2}\theta \sec^{3}\theta d\theta
[/tex]

as the x2 term outside the square root will give tan2.

Replace tan2θ with sec2θ -1 and try to simplify it. I think the integrals may get a bit complicated.
Hi Bazzinga.

The integral:
[tex] 16 \int \tan^{3}\theta \sec^{3}\theta \,d\theta [/tex]
works out very easily with rock.freak667's suggestion for a replacement.

However, if rock.freak667 is right (probably the case) then your integral reduces to
[tex]
16 \int \sec^{5}\theta -\sec^{3}\theta d\theta
[/tex]
 
  • #5
Oh sorry, the term is [tex]x^{3}[/tex] not [tex]x^{2}[/tex]
 
  • #6
Bazzinga said:
Oh sorry, the term is [tex]x^{3}[/tex] not [tex]x^{2}[/tex]
In that case,
[tex]
\int \tan^{3}\theta \sec^{3}\theta \,d\theta = \int \tan\theta\,\left(\sec^{5}\theta-\sec^{3}\theta \right)\, d\theta
[/tex]
[tex]=\int \left(\sec^{4}\theta-\sec^{2}\theta \right)\tan\theta\,\sec\theta\ d\theta[/tex]
This is quite straight forward.
 

Related to Inverse Trig Substitution Integral

What is the concept of Inverse Trig Substitution Integral?

The Inverse Trig Substitution Integral is a method used to solve integrals that involve functions with trigonometric expressions in the form of sin, cos, tan, or their inverse functions. It involves substituting a variable with a trigonometric expression to simplify the integral.

When is Inverse Trig Substitution Integral used?

This method is used when the integral involves a square root of a quadratic expression, or when the integral involves a combination of trigonometric functions and algebraic functions.

How do I determine which trigonometric expression to substitute?

The substitution is chosen based on the form of the integral. For example, if the integral involves a square root of a quadratic expression, the substitution x = asin(t) or x = acos(t) can be used. If the integral involves a quadratic expression without a square root, the substitution x = atan(t) can be used.

What are some common mistakes to avoid when using Inverse Trig Substitution Integral?

One common mistake is forgetting to substitute for the differential dx when using the substitution. Another mistake is not correctly choosing the trigonometric expression to substitute, which can result in an incorrect solution.

Can Inverse Trig Substitution Integral be used for all integrals involving trigonometric functions?

No, this method can only be used for integrals that involve a combination of trigonometric and algebraic functions. It cannot be used for integrals that involve only trigonometric functions or only algebraic functions.

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