Solving an Integral Problem with Trig Substitution

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In summary, the conversation is about a student solving a calculus problem using trig substitution. They use the substitution x = 7secθ and end up with the integral 7∫(sec^2(θ) - secθ) dθ. After putting x back into the equation, they get √(x^2-49) - 7 ln|(x/7) + (√(x^2-49)/7)| + C. The student is unsure about their answer and asks for help. Another person suggests rearranging the answer by using the property log(a/b) = log(a) - log(b).
  • #1
Temp0
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Homework Statement



http://i.imgur.com/u1De0i3.png

Homework Equations





The Attempt at a Solution



So I notice that the bottom is in the form x^2 - a^2 where a = 7, so I use trig substitution to start this off.

x = 7secθ, dx = 7secθtanθ, and finally, x^2 - 49 = 49sec(θ)^2 - 49 = 49 tan^2(θ)
Substituting into the integral, I get
7∫(sec^2(θ) - secθ) dθ, which basically turns into:
7tanθ - 7 ln |secθ + tanθ|.
After putting x back into the equation, I end up with:
√(x^2-49) - 7 ln|(x/7) + (√(x^2-49)/7)| + C
I would just like your help in checking my answers, because I don't get any of the answers provided in the multiple choice, and I'm always hesitant to pick "none of the above". Thank you.
 
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  • #2
Temp0 said:

Homework Statement



http://i.imgur.com/u1De0i3.png

Homework Equations


The Attempt at a Solution



So I notice that the bottom is in the form x^2 - a^2 where a = 7, so I use trig substitution to start this off.

x = 7secθ, dx = 7secθtanθ, and finally, x^2 - 49 = 49sec(θ)^2 - 49 = 49 tan^2(θ)
Substituting into the integral, I get
7∫(sec^2(θ) - secθ) dθ, which basically turns into:
7tanθ - 7 ln |secθ + tanθ|.
After putting x back into the equation, I end up with:
√(x^2-49) - 7 ln|(x/7) + (√(x^2-49)/7)| + C
I would just like your help in checking my answers, because I don't get any of the answers provided in the multiple choice, and I'm always hesitant to pick "none of the above". Thank you.

One of those answers differs from your answer by a constant.
 
  • #3
What do you mean? Hmm, I can't really see any way to rearrange it like that, I think I know the one you're talking about though.
 
  • #4
Temp0 said:
What do you mean? Hmm, I can't really see any way to rearrange it like that, I think I know the one you're talking about though.

For example, log(x/7)=log(x)-log(7).
 
  • #5
Ohhhh! Let's see if I can go any further now, thanks alot.
 

1. What is "Another Integral Problem"?

"Another Integral Problem" refers to a mathematical question that involves finding the integral or antiderivative of a given function. It is a fundamental concept in calculus and is used to solve various real-world problems in physics, engineering, and economics.

2. How is "Another Integral Problem" different from other types of integrals?

"Another Integral Problem" is a specific type of integral that often involves more complex functions or requires advanced techniques to solve. It may also involve multiple variables or require the use of special formulas or substitutions.

3. What methods can be used to solve "Another Integral Problem"?

There are several methods that can be used to solve "Another Integral Problem", including substitution, integration by parts, trigonometric substitution, and partial fractions. The method used will depend on the complexity of the function and the specific problem.

4. How is "Another Integral Problem" related to derivatives?

"Another Integral Problem" and derivatives are inversely related. The derivative of a function represents its rate of change, while the integral represents the accumulation of the function over an interval. In other words, the integral is the opposite operation of taking a derivative.

5. What are some real-world applications of "Another Integral Problem"?

"Another Integral Problem" has many real-world applications, including calculating distances and areas, determining the volume of irregular shapes, finding the work done by a varying force, and predicting future values in economics and finance.

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