Solving A Complex Integral: A Step-by-Step Guide

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Unraveling the Mystery: Solving the Integral of ∫ [(3t² - 1) / √(t(t² - 1))] arcsin((t+1)/(t-1)) dt

Hey everyone! Today, we're diving deep into a fascinating integral that looks intimidating at first glance. Our mission? To solve:

I=∫3t2−1t(t2−1)Arcsen(t+1t−1)dtI = \int \frac{3t^2 - 1}{\sqrt{t(t^2 - 1)}} \text{Arcsen}\left(\frac{t+1}{t-1} \right) dt

Buckle up, because this is going to be a thrilling ride through the world of calculus! We'll break it down step by step, making sure everyone can follow along. So, let's get started and demystify this complex integral together!

Initial Assessment: Recognizing the Challenge

Alright, before we jump into solving, let's take a good look at what we're dealing with. The integral involves a few key components that make it particularly interesting:

  • A rational function: (3t² - 1) / √(t(t² - 1))
  • An inverse trigonometric function: arcsin((t+1)/(t-1))

The combination of these elements suggests that we might need to employ a mix of techniques, possibly including substitution, integration by parts, and trigonometric identities. The presence of the square root and the arcsine function are strong hints that strategic substitutions will be crucial. Our main keyword here is the complex integral, which we aim to simplify. First, note that the expression inside the square root, t(t² - 1), and the argument of the arcsine function, (t+1)/(t-1), both suggest potential algebraic manipulations. Let's keep these in mind as we move forward. Recognizing these challenges is the first step towards finding a solution, guys! Remember, the more complex the problem, the more satisfying the solution. So, let’s roll up our sleeves and get into the nitty-gritty of how to tackle this integral. Understanding the problem thoroughly is half the battle won, and we're already on our way.

Strategic Substitution: Simplifying the Integral

Okay, so the integral looks pretty intense, right? Let’s try a clever substitution to make things a bit easier. Notice that the term inside the square root is t(t² - 1) = t³ - t. Also, we have an arcsin function with (t+1)/(t-1) as its argument. A useful substitution might be related to simplifying the square root. Let's try substituting:

u=t3−tu = \sqrt{t^3 - t}

Squaring both sides, we get:

u2=t3−tu^2 = t^3 - t

Now, let's differentiate both sides with respect to t:

2ududt=3t2−12u \frac{du}{dt} = 3t^2 - 1

dudt=3t2−12u=3t2−12t3−t\frac{du}{dt} = \frac{3t^2 - 1}{2u} = \frac{3t^2 - 1}{2\sqrt{t^3 - t}}

So, $du = \frac{3t^2 - 1}{2\sqrt{t^3 - t}} dt$

Notice that our integral has the term $\frac{3t^2 - 1}{\sqrt{t(t^2 - 1)}} dt$ which is very similar to what we have in our du. Specifically,

3t2−1t(t2−1)dt=2du\frac{3t^2 - 1}{\sqrt{t(t^2 - 1)}} dt = 2 du

Now we can rewrite our integral in terms of u:

I=∫2Arcsen(t+1t−1)duI = \int 2 \text{Arcsen}\left(\frac{t+1}{t-1}\right) du

This looks a bit simpler, doesn't it? But we still have that arcsin term with t in it. We need to express $\text{Arcsen}\left(\frac{t+1}{t-1}\right)$ in terms of u. This is where things get a little tricky, and we might need another substitution or a clever manipulation. Keep in mind our main keywords: integral simplification. We've successfully simplified part of the integral, but we’re not quite there yet. Hang in there, guys! We're making progress, and sometimes the trickiest problems require a few layers of substitutions to peel back. Let's keep digging and see if we can find a way to express the arcsin term in terms of u.

Integration by Parts: Tackling the Arcsin Term

Alright, so we've managed to simplify the integral to:

I=∫2Arcsen(t+1t−1)duI = \int 2 \text{Arcsen}\left(\frac{t+1}{t-1}\right) du

Now we need to deal with that pesky arcsin term. It doesn't seem like we can easily express $\text{Arcsen}\left(\frac{t+1}{t-1}\right)$ directly in terms of u. This suggests that we might want to use integration by parts. Recall that integration by parts is given by:

∫vdw=vw−∫wdv\int v dw = vw - \int w dv

Let's choose:

v=2Arcsen(t+1t−1)v = 2\text{Arcsen}\left(\frac{t+1}{t-1}\right)

dw=dudw = du

Then we have:

dv=2ddt[Arcsen(t+1t−1)]dtdv = 2 \frac{d}{dt} \left[ \text{Arcsen}\left(\frac{t+1}{t-1}\right) \right] dt

w=u=t3−tw = u = \sqrt{t^3 - t}

Now we need to find the derivative of the arcsin term:

ddt[Arcsen(t+1t−1)]=11−(t+1t−1)2⋅ddt(t+1t−1)\frac{d}{dt} \left[ \text{Arcsen}\left(\frac{t+1}{t-1}\right) \right] = \frac{1}{\sqrt{1 - \left(\frac{t+1}{t-1}\right)^2}} \cdot \frac{d}{dt} \left(\frac{t+1}{t-1}\right)

First, let's find the derivative of (t+1)/(t-1):

ddt(t+1t−1)=(t−1)(1)−(t+1)(1)(t−1)2=t−1−t−1(t−1)2=−2(t−1)2\frac{d}{dt} \left(\frac{t+1}{t-1}\right) = \frac{(t-1)(1) - (t+1)(1)}{(t-1)^2} = \frac{t - 1 - t - 1}{(t-1)^2} = \frac{-2}{(t-1)^2}

Now, let's simplify the term inside the square root:

1−(t+1t−1)2=1−(t+1)2(t−1)2=(t−1)2−(t+1)2(t−1)2=t2−2t+1−(t2+2t+1)(t−1)2=−4t(t−1)21 - \left(\frac{t+1}{t-1}\right)^2 = 1 - \frac{(t+1)^2}{(t-1)^2} = \frac{(t-1)^2 - (t+1)^2}{(t-1)^2} = \frac{t^2 - 2t + 1 - (t^2 + 2t + 1)}{(t-1)^2} = \frac{-4t}{(t-1)^2}

So,

11−(t+1t−1)2=1−4t(t−1)2=∣t−1∣−4t=∣t−1∣2it\frac{1}{\sqrt{1 - \left(\frac{t+1}{t-1}\right)^2}} = \frac{1}{\sqrt{\frac{-4t}{(t-1)^2}}} = \frac{|t-1|}{\sqrt{-4t}} = \frac{|t-1|}{2i\sqrt{t}}

(Assuming t<0). Thus,

dv=2⋅∣t−1∣2it⋅−2(t−1)2dt=−2∣t−1∣it(t−1)2dtdv = 2 \cdot \frac{|t-1|}{2i\sqrt{t}} \cdot \frac{-2}{(t-1)^2} dt = \frac{-2|t-1|}{i\sqrt{t} (t-1)^2} dt

Now we can apply integration by parts:

I=2Arcsen(t+1t−1)t3−t−∫t3−t⋅−2∣t−1∣it(t−1)2dtI = 2\text{Arcsen}\left(\frac{t+1}{t-1}\right) \sqrt{t^3 - t} - \int \sqrt{t^3 - t} \cdot \frac{-2|t-1|}{i\sqrt{t} (t-1)^2} dt

I=2Arcsen(t+1t−1)t3−t+∫t3−t⋅2∣t−1∣it(t−1)2dtI = 2\text{Arcsen}\left(\frac{t+1}{t-1}\right) \sqrt{t^3 - t} + \int \sqrt{t^3 - t} \cdot \frac{2|t-1|}{i\sqrt{t} (t-1)^2} dt

This looks even more complicated, doesn’t it? But don’t worry, sometimes integration by parts leads to a more manageable form after simplification. The main keyword here remains integral simplification. We've successfully applied integration by parts, but we're not quite at the finish line yet. Let's see if we can simplify the resulting integral further and get closer to a solution. Remember, guys, complex integrals often require multiple steps and a bit of algebraic gymnastics.

Further Simplification: Algebraic Manipulations

Okay, let's see if we can simplify that integral we got after applying integration by parts:

I=2Arcsen(t+1t−1)t3−t+∫t3−t⋅2∣t−1∣it(t−1)2dtI = 2\text{Arcsen}\left(\frac{t+1}{t-1}\right) \sqrt{t^3 - t} + \int \sqrt{t^3 - t} \cdot \frac{2|t-1|}{i\sqrt{t} (t-1)^2} dt

We can rewrite the integral as:

I=2Arcsen(t+1t−1)t3−t+2i∫t(t2−1)∣t−1∣t(t−1)2dtI = 2\text{Arcsen}\left(\frac{t+1}{t-1}\right) \sqrt{t^3 - t} + \frac{2}{i} \int \frac{\sqrt{t(t^2 - 1)} |t-1|}{\sqrt{t} (t-1)^2} dt

I=2Arcsen(t+1t−1)t3−t+2i∫t2−1∣t−1∣(t−1)2dtI = 2\text{Arcsen}\left(\frac{t+1}{t-1}\right) \sqrt{t^3 - t} + \frac{2}{i} \int \frac{\sqrt{t^2 - 1} |t-1|}{(t-1)^2} dt

Now, let's consider the case when t > 1, so |t-1| = t-1. Then:

I=2Arcsen(t+1t−1)t3−t+2i∫t2−1t−1dtI = 2\text{Arcsen}\left(\frac{t+1}{t-1}\right) \sqrt{t^3 - t} + \frac{2}{i} \int \frac{\sqrt{t^2 - 1}}{t-1} dt

Now we can try another substitution. Let's try:

t=cosh⁡(x)t = \cosh(x)

Then $dt = \sinh(x) dx$

So, $\sqrt{t^2 - 1} = \sqrt{\cosh^2(x) - 1} = \sqrt{\sinh^2(x)} = \sinh(x)$

And $t - 1 = \cosh(x) - 1$

Now the integral becomes:

∫t2−1t−1dt=∫sinh⁡(x)cosh⁡(x)−1sinh⁡(x)dx=∫sinh⁡2(x)cosh⁡(x)−1dx\int \frac{\sqrt{t^2 - 1}}{t-1} dt = \int \frac{\sinh(x)}{\cosh(x) - 1} \sinh(x) dx = \int \frac{\sinh^2(x)}{\cosh(x) - 1} dx

Using the identity $\sinh^2(x) = \cosh^2(x) - 1$, we get:

∫cosh⁡2(x)−1cosh⁡(x)−1dx=∫(cosh⁡(x)−1)(cosh⁡(x)+1)cosh⁡(x)−1dx=∫(cosh⁡(x)+1)dx\int \frac{\cosh^2(x) - 1}{\cosh(x) - 1} dx = \int \frac{(\cosh(x) - 1)(\cosh(x) + 1)}{\cosh(x) - 1} dx = \int (\cosh(x) + 1) dx

∫(cosh⁡(x)+1)dx=sinh⁡(x)+x+C\int (\cosh(x) + 1) dx = \sinh(x) + x + C

Now we need to convert back to t. Since $t = \cosh(x)$, we have $x = \text{arccosh}(t)$

And $\sinh(x) = \sqrt{\cosh^2(x) - 1} = \sqrt{t^2 - 1}$

So,

∫t2−1t−1dt=t2−1+arccosh(t)+C\int \frac{\sqrt{t^2 - 1}}{t-1} dt = \sqrt{t^2 - 1} + \text{arccosh}(t) + C

Finally, we can plug this back into our original integral:

I=2Arcsen(t+1t−1)t3−t+2i(t2−1+arccosh(t))+CI = 2\text{Arcsen}\left(\frac{t+1}{t-1}\right) \sqrt{t^3 - t} + \frac{2}{i} (\sqrt{t^2 - 1} + \text{arccosh}(t)) + C

This is a complex result, and we should remember that we assumed t > 1. If t < 1, the absolute value |t-1| would change the sign, leading to a different result. The main keyword throughout has been integral simplification. We've made significant progress by using a series of substitutions, integration by parts, and algebraic manipulations. Remember guys, even if the solution looks a bit intimidating, the journey of solving it is what truly matters!

Conclusion: Reflecting on the Solution

So, after all that hard work, we've arrived at a solution for the integral:

I=∫3t2−1t(t2−1)Arcsen(t+1t−1)dtI = \int \frac{3t^2 - 1}{\sqrt{t(t^2 - 1)}} \text{Arcsen}\left(\frac{t+1}{t-1}\right) dt

Our solution, under the assumption that t > 1, is:

I=2Arcsen(t+1t−1)t3−t+2i(t2−1+arccosh(t))+CI = 2\text{Arcsen}\left(\frac{t+1}{t-1}\right) \sqrt{t^3 - t} + \frac{2}{i} (\sqrt{t^2 - 1} + \text{arccosh}(t)) + C

It's important to remember that this solution is valid for t > 1. If t < 1, the absolute value in our calculations would change, leading to a different result. This highlights the importance of paying attention to the conditions under which our substitutions and simplifications are valid. Throughout this process, our guiding principle has been integral simplification, using various techniques like substitution, integration by parts, and algebraic manipulation. Solving complex integrals like this one is a testament to the power of calculus and the beauty of mathematical problem-solving. Even though the final answer might look a bit daunting, the journey to get there involves a lot of interesting techniques and insights. Remember guys, practice makes perfect! The more you tackle these kinds of problems, the more comfortable you'll become with the various techniques involved. And always remember to double-check your assumptions and conditions to ensure your solution is valid.

So there you have it! We've successfully navigated through a complex integral, and hopefully, you've gained a better understanding of the techniques involved. Keep practicing, keep exploring, and keep pushing the boundaries of your mathematical knowledge! Cheers!