Trig substitution help (easy one )

In summary: Cute, you have been a great help!In summary, we can use trig substitution to solve the integral \int_{0}^{5} \frac{dt}{25 + x^2}dt. By substituting x = 5\tan \theta, we can simplify the integral to \int_{0}^{\frac{\pi}{4}} \frac{5\sec^2 \theta}{(25 + 25\tan^2 \theta)^2} \ d\theta. Factoring the denominator and using a power reducing formula, we can further simplify to \int_{0}^{\frac{\pi}{4}} \frac{\cos^2 \theta}{25} d\theta.
  • #1
RadiationX
256
0
Use trig substitution to find [tex]\int_{0}^{5} \frac{dt}{25 + x^2}dt[/tex]

I can solve it to here [tex]\int_{0}^{\frac{\pi}{4}}\frac{25sec^2\theta}{(25 + tan^2\theta)^2}[/tex]

and from this point i can factor the denominator into [tex]{625(1+ \tan^2\theta)}^2[/tex]

which becomes [tex]625\sec^4\theta[/tex]

now i have the integral [tex]\int_{0}^{\frac{\pi}{4}}\frac{25sec^2\theta}{625\sec^4\theta}[/tex]

this now reduces to [tex]\int_{0}^{\frac{\pi}{4}}\frac{cos^2\theta}{25}[/tex]

and at this point i can use a power reducing formula to get rid of the [tex]\cos^2\theta[/tex]

assuming that the last integral is correct and that i use the power reducing formula to reduce [tex]\cos^2\theta[/tex] correctly, what am i doing wrong?
i have a TI-89 graphing calculator, and when i integrat this problem on it i get a different answer than when i do it by hand. where is my mistake?




this post is incorrect look further down for the correction.
i'm really sorry about this.
 
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  • #2
You can use the double angle formula for cos 2x to get rid of cos^2 x. cos 2x = 2cos^2 x - 1.
 
  • #3
As the integral stands now, you could use another formula to find the answer, but in order to put the integral into a form where you can use a trig substution you have to play around with it:

[tex]25 + t^2 = (\sqrt{25 + t^2})^2[/tex]

So the integral becomes:

[tex]\int_{0}^{5}\frac{dt}{(\sqrt{25 + t^2})^2}[/tex]

Make the substutions:
[tex]t = 5 \tan{\theta}[/tex]
[tex]dt = 5 \sec^2{\theta} d\theta[/tex]

and work it from there. :)
 
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  • #4
This is actually a standard integral. Anyways, you seem to have made a few errors. Here we want

[tex] \int_0^5 \frac{dt}{25 + t^2}[/tex]

let [itex] t = 5\tan{\theta} \Longrightarrow dt = 5 \sec^2{\theta}[/itex] and note that [itex]t=5 \Longrightarrow 5 = 5\tan{\theta} \Longrightarrow \theta = \pi / 4[/itex]. Thus we just have

[tex] \int_0^{\frac{\pi}{4}} \frac{5\sec^2 \theta}{25 + (5 \tan \theta)^2} \ d\theta[/tex]

[tex] = \frac{1}{5}\int_0^{\frac{\pi}{4}} \frac{\sec^2 \theta}{1 + \tan^2 \theta} \ d\theta[/tex]

now, review your trig identities and see if you can find an easy way to do this :smile:

The standard integral is

[tex]\int \frac{dx}{a^2+x^2} = \frac{1}{a}\arctan\left(\frac{x}{a}\right) \ + \ C,[/tex]

by the way.
 
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  • #5
RadiationX said:
Use trig substitution to find [tex]\int_{0}^{5} \frac{dt}{25 + x^2}dt[/tex]

I can solve it to here [tex]\int_{0}^{\frac{\pi}{4}}\frac{25sec^2\theta}{(25 + tan^2\theta)^2}[/tex]

and from this point i can factor the denominator into [tex]{625(1+ \tan^2\theta)}^2[/tex]

which becomes [tex]625\sec^4\theta[/tex]

now i have the integral [tex]\int_{0}^{\frac{\pi}{4}}\frac{25sec^2\theta}{625\sec^4\theta}[/tex]

this now reduces to [tex]\int_{0}^{\frac{\pi}{4}}\frac{cos^2\theta}{25}[/tex]

and at this point i can use a power reducing formula to get rid of the [tex]\cos^2\theta[/tex]

assuming that the last integral is correct and that i use the power reducing formula to reduce [tex]\cos^2\theta[/tex] correctly, what am i doing wrong?
i have a TI-89 graphing calculator, and when i integrat this problem on it i get a different answer than when i do it by hand. where is my mistake?






I made a huge mistake in my original post! the first integral is

[tex]\int_{0}^{5} \frac{dt}{(25 + x^2)^2}dt[/tex]

and what follows is what i get. I'm really sorry about posting this problem incorrectly.
 
  • #6
i'm putting this back to the top because i screwed it up the first time.
 
  • #7
No problem :smile:

Anyways, in that case your work is almost right. We want

[tex] \int_0^5 \frac{dx}{(25+x^2)^2}[/tex]

substitute [itex]x = 5\tan \theta \Longrightarrow dx = 5\sec^2 \theta \ d\theta[/itex] and so [itex]x=5 \Longrightarrow \theta = \pi / 4[/itex]. Thus our integral is the same as

[tex] \int_0^{\frac{\pi}{4}} \frac{5\sec^2 \theta}{(25 + 25\tan^2 \theta)^2} \ d\theta[/tex]

[tex] = \frac{1}{125}\int_0^{\frac{\pi}{4}} \frac{\sec^2 \theta}{\sec^4 \theta} \ d\theta[/tex]

[tex] = \frac{1}{125}\int_0^{\frac{\pi}{4}} \cos^2 \theta \ d\theta[/tex]
 
  • #8
Data said:
No problem :smile:

Anyways, in that case your work is almost right. We want

[tex] \int_0^5 \frac{dx}{(25+x^2)^2}[/tex]

substitute [itex]x = 5\tan \theta \Longrightarrow dx = 5\sec^2 \theta \ d\theta[/itex] and so [itex]x=5 \Longrightarrow \theta = \pi / 4[/itex]. Thus our integral is the same as

[tex] \int_0^{\frac{\pi}{4}} \frac{5\sec^2 \theta}{(25 + 25\tan^2 \theta)^2} \ d\theta[/tex]

[tex] = \frac{1}{125}\int_0^{\frac{\pi}{4}} \frac{\sec^2 \theta}{\sec^4 \theta} \ d\theta[/tex]

[tex] = \frac{1}{125}\int_0^{\frac{\pi}{4}} \cos^2 \theta \ d\theta[/tex]

i worked this down to exactly the same thing that you did but when i use my
TI-89 graphing calculator to integrate this i get a different answer than what i get by hand.
 
  • #9
[tex]\frac{1}{125}\int_0^{\frac{\pi}{4}} \cos^2 \theta d\theta[/tex]

[tex] = \frac{1}{125} \int_0^{\frac{\pi}{4}} \frac{1}{2} \left(1 + \cos (2\theta)) \ d\theta[/tex]

[tex] = \frac{1}{250}\left[\theta + \frac{\sin (2\theta)}{2} \right]_0^{\frac{\pi}{4}}[/tex]

[tex] = \frac{1}{250}\left(\frac{\pi}{4} + \frac{1}{2}\right).[/tex]
 
  • #10
now i see my mistake. thankyou Data
 

Related to Trig substitution help (easy one )

1. What is trig substitution?

Trig substitution is a method used in calculus to evaluate integrals involving expressions with trigonometric functions. It involves substituting a trigonometric expression for a variable in the integral to make the integral easier to solve.

2. When should I use trig substitution?

Trig substitution should be used when the integral involves expressions with trigonometric functions, such as sine, cosine, or tangent. It is also helpful when the integral involves expressions with radicals.

3. How do I choose the appropriate trig substitution?

The appropriate trig substitution is determined by the form of the integral. For example, if the integral involves the expression √(a^2 - x^2), you would use the substitution x = a sinθ. If the integral involves the expression √(a^2 + x^2), you would use the substitution x = a tanθ.

4. What are the common trig substitutions used?

The most common trig substitutions used are:

  • x = a sinθ for an expression of the form √(a^2 - x^2)
  • x = a cosθ for an expression of the form √(a^2 + x^2)
  • x = a tanθ for an expression of the form √(a^2 + x^2)

5. Can you provide an example of solving an integral using trig substitution?

Sure, for the integral ∫(x^2)/(√(4 - x^2)) dx, we can use the substitution x = 2 sinθ. This will simplify the integral to ∫(4 sin^2θ)/(2 cosθ) dθ. Using trigonometric identities, we can further simplify this to ∫2 sinθ dθ. Solving this integral gives us -2 cosθ + C. Substituting back for x, we get the final answer of -2√(4 - x^2) + C.

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