Improper integral help (trig sub?)

In summary, the person is trying to solve a homework problem but keeps making mistakes. They substitute the correct answer, pi, for -2sin(x) and get the answer as -2. They realize their mistake and substitute -2sin(x) for pi again to get the answer as pi.
I have been bashing my head against this problem for a couple of hours now and cannot for the life of me figure it out. i am able to get AN answer but when i check it with my calculator i always get pi

Homework Statement

integral from (0, 1) of:

(4r*dr)/sqrt(1 - r^4)

The Attempt at a Solution

i know the equation is undefined at r = 1 so:

lim t --> 1 of the integral (0, t) of (4r*dr)/sqrt(1 - r^4)

by trig subsitution:
sin(x) = sqrt(1-r^4)
cos(x) = r^2
deriving implicitly gives:
-sin(x) = 2rdr
-2sin(x) = 4rdr

so i end up with:
integral (0, t) of (-2sin(x))/(sin(x))

which gives me the answer as -2

maybe my calculator is wrong? or more likely I am making some stupid mistake/overlooking something.

When you do a variable substitution you need to change the limits of integration as well.
So you need to integrate from the x-value when $$\sin(x) = 1$$ to where $$\sin(x) = 0$$ since when $$r = 0, \quad \sqrt{1-r^4} = 1$$ and when $$r = 1, \quad \sqrt{1-r^4}= 0$$.

$$\pi$$ is the correct answer.

Last edited:
try r = [cos(θ)]^1/2

also with your sub i get r^4 = cos(x)

JonF said:
try r = [cos(θ)]^1/2

also with your sub i get r^4 = cos(x)

His substitution is fine. He just forgot to change the limits.
Also $$\sin(x) = \sqrt{1-\cos^2 (x)}$$ so in his case $$r^4 = \cos^2 (x) \Rightarrow r^2 = \cos(x)$$

Inferior89 said:
When you do a variable substitution you need to change the limits of integration as well.
$$\pi$$ is the correct answer. So you need to integrate from the x-value when sin(x) = 1 to where sin(x) = 0 since when r = 0 sqrt(1-r^4) = 1 and when r = 1 sqrt (1-r^4) = 0.

yup, youre right i was making a stupid mistake, i just totally left out the last step.

i just plugged arccos(r^2) in for theta after integrating and got pi as the correct answer. thanks for all of your help.

Inferior89 said:
His substitution is fine. He just forgot to change the limits.
Also $$\sin(x) = \sqrt{1-\cos^2 (x)}$$ so in his case $$r^4 = \cos^2 (x) \Rightarrow r^2 = \cos(x)$$

Was just pointing out there is an easier sub where he may have not run into arithmetic mistakes. Like his claim: sin(x) = sqrt(1-r^4) -> cos(x) = r^2

JonF said:
Was just pointing out there is an easier sub where he may have not run into arithmetic mistakes. Like his claim: sin(x) = sqrt(1-r^4) -> cos(x) = r^2

What is wrong with that claim? Looks right to me.Edit:
Think a right angled triangle with hypotenuse 1 and the opposite side of the angle x as sqrt(1-r^4).
This means that sin(x) = sqrt(1-r^4).
From Pythagoras the last side (y) must satisfy sqrt(1-r^4)^2 + y^2 = 1^2 --> 1 - r^4 + y^2 = 1 --> r^4 = y^2 --> y = r^2.
Now cos(x) = y/1 = r^2.

Last edited:
ah oops i was doing sin(x) = sqrt(1-r^4) ⇒ sin^2(X) = 1 – r^4 ⇒ r^4 = sin^2(x) – 1 ⇒ r^4 = cos^2(x)

and just noticed that he didn't have a ^2 on his cos(x) my bad.

What is an improper integral?

An improper integral is an integral where one or both of the limits of integration is infinite or the integrand function is not defined at some points within the limits of integration.

Why do we need to use trigonometric substitution for improper integrals?

Trigonometric substitution is used to transform an integral with algebraic or rational expressions into one that can be evaluated using trigonometric identities. This is necessary for improper integrals because it allows us to manipulate the integrand into a form that can be evaluated at infinity or at points where it is undefined.

How do you choose the appropriate trigonometric substitution for an improper integral?

The choice of trigonometric substitution depends on the form of the integrand. For a rational function, use the substitution $x = \tan \theta$. For a radical function, use the substitution $x = \sin \theta$ or $x = \cos \theta$. For expressions involving $\sqrt{a^2 - x^2}$, use the substitution $x = a \sin \theta$ or $x = a \cos \theta$.

Can I use any trigonometric substitution for any improper integral?

No, the appropriate substitution must be chosen based on the form of the integrand. Using the wrong substitution may lead to an incorrect answer or make the integral more difficult to evaluate.

What should I do if the limits of integration are not specified for an improper integral?

If the limits of integration are not specified, you should evaluate the integral as a definite integral with variable limits. Once you have a general solution, you can then evaluate the limit as the appropriate limit of integration approaches infinity or a point where the integrand is undefined.

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