Help evaluating these integrals

  • Thread starter Ed Quanta
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    Integrals
In summary, an integral is a mathematical concept used to find the total value of a function over a certain interval. The process for evaluating integrals involves using techniques such as substitution, integration by parts, or trigonometric substitution. Evaluating integrals is important as it allows us to find the exact value of a function, and there are various techniques, including substitution, integration by parts, trigonometric substitution, and partial fraction decomposition. Integrals can have multiple solutions depending on the interval and function, so it is important to check for any restrictions or special cases when evaluating them.
  • #1
Ed Quanta
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Sorry I don't know how to use the notation, but help me solve the following through the u substitution.

1) dx/square root of x^2+5x

2)dx/x^2*square root of 1-x^2

I am embarassed that I am not able to solve these still after devoting much thought, but help from anyone will be appreciated.
 
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  • #2
(1) Complete the square

(2) Let u = 1/x
 
  • #3


No need to be embarrassed, integration can be tricky and it takes practice to become comfortable with it. Let's go through each integral step by step:

1) dx/square root of x^2+5x

To solve this integral, we will use the u-substitution method. Let u = x^2 + 5x, then du = (2x + 5)dx. We can rearrange this to get dx = du/(2x + 5).

Substituting this into the original integral, we get:

∫dx/square root of x^2+5x = ∫du/(2x + 5)

Next, we need to get rid of the x in the denominator. We can do this by factoring out an x from the denominator, giving us:

∫du/(2x + 5) = ∫du/x(2 + 5/x)

Now, we can use the u-substitution again, this time with v = 2 + 5/x. This gives us dv = -5/x^2 dx. Rearranging this, we get dx = -5/(v^2 - 5)dv.

Substituting this into the integral, we get:

∫du/x(2 + 5/x) = ∫-5/(v^2 - 5)dv

Now, we can use a trigonometric substitution to solve this integral. Let v = √5secθ, then dv = √5tanθsecθ dθ. Substituting this into the integral, we get:

∫-5/(v^2 - 5)dv = ∫-5/(5sec^2θ - 5)√5tanθsecθ dθ

Simplifying this, we get:

∫-5/(v^2 - 5)dv = ∫-√5tanθ dθ

Using the trigonometric identity tan^2θ + 1 = sec^2θ, we can rearrange this to get:

∫-√5tanθ dθ = ∫-√5(tan^2θ + 1 - 1) dθ

= ∫-√5(sec^2θ - 1) dθ = ∫-√5(v^2/5 - 1) dθ

= ∫-
 

What is an integral?

An integral is a mathematical concept that represents the area under a curve in a graph. It is used to find the total value of a function over a certain interval.

What is the process for evaluating integrals?

The process for evaluating integrals involves using techniques such as substitution, integration by parts, or trigonometric substitution to simplify the integral and find its solution.

Why is it important to evaluate integrals?

Evaluating integrals is important because it allows us to find the exact value of a function over a certain interval. This can be useful in a variety of fields, such as physics, engineering, and economics.

What are some common techniques for evaluating integrals?

Some common techniques for evaluating integrals include substitution, integration by parts, trigonometric substitution, and partial fraction decomposition.

Can integrals have multiple solutions?

Yes, integrals can have multiple solutions depending on the interval and the function being integrated. It is important to check for any restrictions or special cases when evaluating integrals.

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