Solving the Integral of √(1 1/(2t))

Integral problems in calculus often present unique challenges, especially when dealing with complex functions such as the square root of a rational expression. In this article, we will delve into the process of integrating the function √(1 1/(2t)).

Introduction to the Problem

The integral in question is the following:

Let I √(1 1/(2t))

I int √(1 1/(2t)) dt

Our goal is to find a solution to this integral by applying various techniques in integral calculus.

Method 1: Substitution and Simplification

We start by simplifying the expression under the square root:

Let t u2, hence dt 2u du

Therefore, the integral transforms into:

Let √(1 1/(2t)) u

This yields: t 1/(2)(1/u2 - 1)

Substituting and simplifying:

The integral becomes: - int 1/(u2 - 1) du

By partial fraction decomposition:

1/(u2 - 1) 1/4(1/(u - 1) - 1/(u 1))

Solving the integral:

I 1/4 int (1/(u - 1) - 1/(u 1)) du

This is a standard form and the solution is:

I 1/4 ln |(u - 1)/(u 1)| C

Substitute back the original variable:

I 1/4 ln |(√(1 1/(2t)) - 1)/(√(1 1/(2t)) 1)| C

Method 2: Another Approach

Alternatively, we can use a different substitution:

Let √(1 1/(2t)) u

Therefore, 1/√(t) 2du

The integral transforms into:

I 2 ∫ u2 (1/2) du

This integral is a standard form and the solution can be found in high school calculus books.

Method 3: Integration by Parts

We can also use integration by parts to solve this integral:

I ∫ √(1 1/(2t) dt)

Let √(1 1/(2t)) u, hence 1/2 √(1 1/(2t)) t

Using the integration by parts formula:

I t √(1 1/(2t)) - 1/4 ∫ t/√(1 1/(2t)) dt

Let 1/(2t) tan2θ, hence 2t cot2θ

Therefore, the integral transforms into:

I t √(1 1/(2t)) 1/2 ln |√(1 1/(2t)) √(2t / (1 1/(2t)))| C

Substitute back u √(1 1/(2t)):

I t √(1 1/(2t)) 1/2 ln |u √(2t / (u2 - 1))| C

Conclusion

By examining various methods, we have successfully solved the integral of √(1 1/(2t)). Each method provides a unique perspective and demonstrates the power of calculus in solving complex problems. The solutions obtained here can be useful for both theoretical understanding and practical applications in various fields of science and engineering.