Integration using Trig Substitution

In summary, the homework statement is trying to find an equation that resembles the original, but with secants and tangents in place of cosine and sine. After substituting y=sin(x), the integrand becomes a standard integral, with sinh^{-1}y + C as the solution.
  • #1
Brickster
21
0

Homework Statement



[tex]\int \frac{cosx dx}{\sqrt{1 + sin^{2}x}}[/tex]

Homework Equations



Expression: [tex]\sqrt{a^{2} + x^{2}}[/tex]
Substitution: x = a*tan[tex]\Theta[/tex]
Identity: 1 + tan[tex]^{2}\Theta = sec^{2}\Theta[/tex]

The Attempt at a Solution


I have tried using Trig Substitution, but I end up getting an equation much like the one I started only it contains secants and tangents instead of cosine and sine. For some reason I only see a circular route that would just take me back to the original equation.
 
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  • #2
Try the substitution y=sin(x). That should then put the integrand in a familiar form.
 
  • #3
as in changing the bottom from 1 + (sinx)^2 to 1 + y^2 ?I can get to:

[tex]\int \frac{sec^{2}\Theta d\Theta}{\sqrt{1+tan^{2}\Theta}}[/tex]

But after that it makes no sense, I just get back to the original equation when I sub. again.
 
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  • #4
After y=sin x, and then a tan theta = y, you should have the same integrand with a plus sign, not a minus. Then use the Pythagorean identities to simplify the square root.
 
  • #5
So I should end up with:

[tex]\int sec\Theta d\Theta}[/tex]

Correct?

then I would get ln [tex]\left| sec\Theta+tan\Theta \right|[/tex] + C
 
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  • #6
If you use the substitution I suggested, then you end up with [tex]\int\frac{dy}{\sqrt{1+y^2}}[/tex]. This is a standard integral, is it not?
 
  • #7
cristo said:
If you use the substitution I suggested, then you end up with [tex]\int\frac{dy}{\sqrt{1+y^2}}[/tex]. This is a standard integral, is it not?

Yes, I'm fairly certain that is a standard integral, and it comes out to:

[tex]sin^{-1}y + C[/tex]

But I'm also fairly certain that I am supposed to solve this problem using the expression, substitution, and identity when I do the trig sub. Maybe I am overcomplicated the problem and should just stick to the arcsin answer, Gib Z's method seems to be the one I'm after
 
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  • #8
Brickster said:
Yes, I'm fairly certain that is a standard integral, and it comes out to:

[tex]sin^{-1}\frac{y} + C[/tex]

Close; the solution is [tex]\sinh^{-1} y +C[/tex]
 
  • #9
cristo said:
Close; the solution is [tex]\sinh^{-1} y +C[/tex]

My mistake, you're right. Thanks!
 
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  • #10
It's always nice to know alternative methods, because we can know express the inverse hyperbolic sine function in terms of the natural logarithm =] Which we could have also done more directly, but o well!
 

Related to Integration using Trig Substitution

What is Trig Substitution?

Trig substitution is a method used in calculus to simplify and solve integrals that involve trigonometric functions.

When should I use Trig Substitution?

Trig substitution is used when the integral contains a combination of algebraic and trigonometric terms, such as square roots or powers of trigonometric functions.

What are the common trigonometric identities used in Trig Substitution?

The most commonly used trigonometric identities in Trig Substitution are sin²x + cos²x = 1, tan²x + 1 = sec²x, and 1 + cot²x = csc²x.

How do I choose the appropriate trigonometric substitution?

The choice of trigonometric substitution depends on the form of the integral. For integrals with √(a²-x²), use x = a sinθ. For integrals with √(a²+x²), use x = a tanθ. And for integrals with √(x²-a²), use x = a secθ.

Are there any tips for solving integrals using Trig Substitution?

One helpful tip is to always check the limits of integration and adjust them accordingly after substituting. It is also important to simplify the integral as much as possible before attempting to integrate it.

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