Solving integral from 5/(x^2 + 1) dx from -1 to 1 works one way but

[s3a]

Homework Statement

The integral of 5/(x^2 + 1) dx from -1 to 1. (The TheIntegral.jpg attachment shows this in a aesthetically-pleasing way.)

Homework Equations

sin^2 (θ) + cos^2 (θ) = 1
1 + tan^2 (θ) = sec^2 (θ)
cot^2 (θ) + 1 = csc^2 (θ)

x = tan(θ)
x = cot(θ)

The Attempt at a Solution

When I am solving the integral, I successfully compute it using the x = tan(θ) substitution but when I use the x = cot(θ) substitution, I get the wrong answer.

My specific work is attached as MyWork.jpg. The work in black is the work I got correct (as indicated by the red writing) and, the work in green is the work I get wrong (as indicated by the red writing).

Could someone please tell me what I am doing wrong when using the cot(θ) substitution?

 

[gopher_p]

$\text{cot}^{-1}1\neq \frac{1}{\tan^{-1}1}$

The inverse trig functions do not satisfy the same identities as the trig functions.

 

[pasmith]

Homework Statement

The integral of 5/(x^2 + 1) dx from -1 to 1. (The TheIntegral.jpg attachment shows this in a aesthetically-pleasing way.)

Homework Equations

sin^2 (θ) + cos^2 (θ) = 1
1 + tan^2 (θ) = sec^2 (θ)
cot^2 (θ) + 1 = csc^2 (θ)

x = tan(θ)
x = cot(θ)

The Attempt at a Solution

When I am solving the integral, I successfully compute it using the x = tan(θ) substitution but when I use the x = cot(θ) substitution, I get the wrong answer.

My specific work is attached as MyWork.jpg. The work in black is the work I got correct (as indicated by the red writing) and, the work in green is the work I get wrong (as indicated by the red writing).

Could someone please tell me what I am doing wrong when using the cot(θ) substitution?

$\cot(\theta)$ blows up when $\tan(\theta) = 0$, ie. when $\theta = n\pi$.

So you need to find values for $\cot^{-1}(\pm 1)$ which lie between $0$ and $\pi$. $\cot^{-1}(1) = \frac{\pi}{4}$ and $\cot^{-1}(-1) = \frac{3\pi}{4}$ will do.

 

[s3a]

Thanks, gother_p, for specifying.

Also, pasmith, thanks for your answer too however, although I can see how using 3π/4 would work algebraically, I don't get how I would know to choose 3π/4 over –π/2; when θ = –π/2, θ ≠ nπ (for n being any integer).

Also, instead of values between 0 and π, I could choose values between 0 and –π if I wanted, right?

 

[s3a]

Actually, I meant θ = –π/4 instead of θ = –π/2.

Also, I'm still confused as to why I can't choose θ = –π/4 instead of θ = 3π/4 (for the reason I listed in my previous post) and, I would really appreciate it if this can be elaborated upon.

Lastly, if I'm not being clear, tell me what to clarify and, I will do so.

 

[eumyang]

Actually, I meant θ = –π/4 instead of θ = –π/2.

Also, I'm still confused as to why I can't choose θ = –π/4 instead of θ = 3π/4 (for the reason I listed in my previous post) and, I would really appreciate it if this can be elaborated upon.

Lastly, if I'm not being clear, tell me what to clarify and, I will do so.

You can't choose $\theta = -\frac{\pi}{4}$, because by definition the range of the inverse cotangent function $\theta = \cot^{-1} x$ is $0 < \theta < \pi$. There is only one value for $\cot^{-1} \left( -1 \right)$ and that is $\frac{3\pi}{4}$.

 

[s3a]

Oh, I think I get it now!

To attempt to restate what you guys said using graphical terminology, is it because in the graph of f(x) = cot(x), there is an asymptote separating –π/4 and π/4 so, we could choose to replace –π/4 with 3π/4, for example, because, the cotangent of –π/4 and 3π/4 yield the the same value or, equivalently, we could choose any other two angles both of which are between the same asymptotes such as –π/4 and –3π/4?

 

[pasmith]

Oh, I think I get it now!

To attempt to restate what you guys said using graphical terminology, is it because in the graph of f(x) = cot(x), there is an asymptote separating –π/4 and π/4 so, we could choose to replace –π/4 with 3π/4, for example, because, the cotangent of –π/4 and 3π/4 yield the the same value or, equivalently, we could choose any other two angles both of which are between the same asymptotes such as –π/4 and –3π/4?

Yes.

 

[s3a]

Yes.

Thanks for confirming. :)