Evaluating Definite Integral: Step-by-Step Solution For $\int_{0}^{\frac{\pi}{2}} \arctan(\sqrt{\tan(x)}+\sqrt{\cot(x)}) Dx$

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Hey guys! Today, we're diving deep into the fascinating world of definite integrals, and we're tackling a particularly juicy one. If you've ever stared blankly at an integral that seems to defy your usual techniques, you're in the right place. We're going to break down the evaluation of this integral step by step, making it crystal clear and, dare I say, even a little fun.

The Integral Challenge

So, what's the integral that's got us all fired up? It's this beauty:

$\qquad I = \int_{0}^{\frac{\pi}{2}} \arctan(\sqrt{\tan(x)}+\sqrt{\cot(x)}) dx$

At first glance, it might seem a bit intimidating. The combination of the arctangent function, square roots, and trigonometric functions can make even seasoned calculus veterans pause for thought. But fear not! We're going to approach this methodically and unlock its secrets. We will explore how to solve this definite integral, which involves a clever substitution and simplification strategy to arrive at the solution. Understanding these steps will not only help in solving this particular integral but also provide insights into tackling similar complex problems.

The Substitution Strategy: Our First Move

The key to cracking this integral lies in a clever substitution. Let's introduce a new variable, $y$, defined as:

$\qquad y^2 = \tan(x)$

This substitution is strategic because it helps us simplify the trigonometric terms within the arctangent function. Now, we need to express everything in terms of $y$, including the differential $dx$. Differentiating both sides of our substitution with respect to $x$, we get:

$\qquad 2y \frac{dy}{dx} = \sec^2(x)$

Remember that $\sec^2(x) = 1 + \tan^2(x)$. Since $y^2 = \tan(x)$, we can rewrite this as:

$\qquad 2y \frac{dy}{dx} = 1 + y^4$

Solving for $dx$, we have:

$\qquad dx = \frac{2y}{1 + y^4} dy$

This is a crucial piece of the puzzle. We've now expressed $dx$ in terms of $dy$, which will allow us to change the variable of integration. It's important to pause and appreciate the elegance of this substitution. By transforming the integral into a different variable space, we are setting the stage for simplification. The initial complexity arising from the mix of trigonometric and inverse trigonometric functions is gradually being addressed through this strategic change of variables. Remember to carefully execute each step of the substitution, ensuring that all components of the original integral, including the limits of integration, are correctly transformed.

Transforming the Integral: Putting it All Together

Now we need to consider the limits of integration. When $x = 0$, we have $\tan(0) = 0$, so $y = 0$. As $x$ approaches $\frac{\pi}{2}$, $\tan(x)$ approaches infinity, so $y$ also approaches infinity. Thus, our integral transforms to:

$\qquad I = \int_{0}^{\infty} \frac{2y}{1 + y^4} \arctan\left( \sqrt{y^2} + \sqrt{\frac{1}{y^2}} \right) dy$

Simplifying the expression inside the arctangent, we get:

$\qquad I = \int_{0}^{\infty} \frac{2y}{1 + y^4} \arctan\left( y + \frac{1}{y} \right) dy$

This looks much cleaner already! The square roots are gone, and we have a more manageable expression inside the arctangent. This transformation is a testament to the power of strategic substitution. By choosing an appropriate substitution, we've converted a seemingly intractable integral into a form that's ripe for further simplification. At this point, the integral is still challenging, but the pathway towards a solution is becoming clearer. The next steps will likely involve further simplification or the application of another clever technique to fully evaluate the integral. Make sure you understand how the limits of integration have changed and the algebraic manipulations that led to the simplified form. This careful attention to detail is essential in ensuring the correctness of the solution.

Further Simplification: Unveiling the Inner Workings

The expression $y + \frac{1}{y}$ inside the arctangent suggests another possible simplification. Let's focus on this term. We can rewrite it as:

$\qquad y + \frac{1}{y} = \frac{y^2 + 1}{y}$

Now, the integral becomes:

$\qquad I = \int_{0}^{\infty} \frac{2y}{1 + y^4} \arctan\left( \frac{y^2 + 1}{y} \right) dy$

This form hints at a possible trigonometric identity we can exploit. Recall the tangent addition formula:

$\qquad \tan(a + b) = \frac{\tan(a) + \tan(b)}{1 - \tan(a)\tan(b)}$

Let's consider the arctangent of a sum:

$\qquad \arctan(a) + \arctan(b) = \arctan\left( \frac{a + b}{1 - ab} \right)$

This identity might be useful in simplifying the arctangent term in our integral. However, it's not a direct fit just yet. We need to manipulate the expression inside the arctangent to match the form of the identity. To effectively apply this identity, we need to carefully analyze the expression $\frac{y^2 + 1}{y}$ and see how it can be related to the $\frac{a + b}{1 - ab}$ structure. This often involves thinking creatively about how to rewrite the numerator and denominator to reveal the underlying structure. Don't be afraid to experiment with different algebraic manipulations. The key is to try to find a form that aligns with known identities or standard integral forms. This step highlights the importance of having a good grasp of trigonometric identities and algebraic manipulation techniques in solving complex integrals.

A Trigonometric Twist: Applying the Arctangent Identity

Let's try to express $\frac{y^2 + 1}{y}$ in a form that allows us to use the arctangent addition identity. We can rewrite the arctangent term as:

$\qquad \arctan\left( \frac{y^2 + 1}{y} \right) = \arctan\left( \frac{y + \frac{1}{y}}{1 - (y)(\frac{1}{y}) + 1} \right)$

This doesn't immediately fit the arctangent addition formula. Let's try something else. Recall that: $\arctan(y)-\arctan(1/y) = \arctan(\frac{y-1/y}{1+y(1/y)}) = \arctan(\frac{y^2-1}{2y})$

Instead, consider using the identity: $\qquad \arctan(y) + \arctan(\frac{1}{y}) = \frac{\pi}{2}, \text{ for } y > 0$

This identity is incredibly useful here! It directly addresses the form we have inside the arctangent. Substituting this into our integral, we get:

$\qquad I = \int_{0}^{\infty} \frac{2y}{1 + y^4} \cdot \frac{\pi}{2} dy$

This is a significant simplification! The arctangent term has vanished, replaced by a constant. Our integral now looks much more manageable. This step beautifully illustrates the power of recognizing and applying appropriate trigonometric identities. The integral has been transformed from a daunting expression to a much simpler form, ready for the final integration. Take a moment to appreciate how the application of a single identity can drastically alter the complexity of a problem. This underscores the importance of building a strong repertoire of mathematical tools and techniques to draw upon when faced with challenging problems.

The Final Integration: Reaching the Solution

Now we have a much simpler integral:

$\qquad I = \pi \int_{0}^{\infty} \frac{y}{1 + y^4} dy$

To evaluate this, let's make another substitution: $u = y^2$. Then, $du = 2y dy$, so $y dy = \frac{1}{2} du$. The limits of integration remain the same (0 to infinity). The integral becomes:

$\qquad I = \frac{\pi}{2} \int_{0}^{\infty} \frac{1}{1 + u^2} du$

This is a standard integral! We know that:

$\qquad \int \frac{1}{1 + u^2} du = \arctan(u) + C$

So, we have:

$\qquad I = \frac{\pi}{2} \left[ \arctan(u) \right]_{0}^{\infty}$

Evaluating the limits:

$\qquad I = \frac{\pi}{2} \left( \frac{\pi}{2} - 0 \right) = \frac{\pi^2}{4}$

And there we have it! The value of the integral is:

$\qquad I = \frac{\pi^2}{4}$

This final integration step brings our journey to a successful conclusion. By carefully applying substitutions and trigonometric identities, we've managed to evaluate a seemingly complex integral. The result, $\frac{\pi^2}{4}$, is a testament to the power of mathematical techniques. Reflect on the entire process, from the initial substitution to the final evaluation. Each step played a crucial role in unraveling the integral. The ability to recognize standard integral forms and apply them correctly is essential in calculus. This final step reinforces the importance of having a solid understanding of basic integration rules and techniques.

Conclusion: A Triumph of Technique

We've successfully navigated this challenging integral, and hopefully, you've gained some valuable insights along the way. The key takeaways are the power of strategic substitutions, the importance of trigonometric identities, and the methodical approach to problem-solving. This journey not only helped us solve a specific integral but also enhanced our problem-solving toolkit for future challenges. So, keep practicing, keep exploring, and keep those integrals coming! Remember, the beauty of mathematics lies in the journey of discovery. Each problem solved is a step forward in your mathematical understanding. This example showcases how seemingly complex problems can be tackled effectively by breaking them down into smaller, manageable steps and applying the appropriate techniques. The satisfaction of arriving at the solution makes the effort worthwhile. So, keep challenging yourself, keep learning, and most importantly, keep enjoying the process of mathematical exploration.

How to evaluate the definite integral $I = \int_{0}^{\frac{\pi}{2}} \arctan(\sqrt{\tan(x)}+\sqrt{\cot(x)}) dx$? Provide a step-by-step solution.

Evaluating Definite Integral: Step-by-Step Solution for $\int_{0}^{\frac{\pi}{2}} \arctan(\sqrt{\tan(x)}+\sqrt{\cot(x)}) dx$