Solving the Integral of ∫√[x/a-x] dx: A Step-by-Step Guide

In calculus, the integration of certain functions often requires advanced techniques such as trigonometric substitutions. This article aims to provide a detailed guide on solving the integral of ∫√[x/a-x] dx. The solution involves a trigonometric substitution, which will be explained step-by-step.

Introduction to the Integral

Let's consider the integral:

I ∫ √[x/a - x] dx

Trigonometric Substitution

To solve this integral, we will use a trigonometric substitution. The first step is to let x a sin2 t, which changes the variable x into a trigonometric function. This substitution is chosen because it simplifies the expression under the square root.

dx 2a sin t cos t dt

Substituting and Simplifying

Substituting x a sin2 t and dx 2a sin t cos t dt into the integral:

I ∫ √[a sin2 t / (a - a sin2 t)] * 2a sin t cos t dt

The denominator simplifies as follows:

∫ √[a sin2 t / (a(1 - sin2 t))] * 2a sin t cos t dt

Simplifying the trigonometric identity 1 - sin2 t cos2 t:

I a ∫ √[sin2 t / cos2 t] * 2 sin t cos t dt a ∫ cos t * 2 sin t dt

The integral now simplifies to:

I a ∫ 1 - cos 2t dt

Evaluating the Integral

Integrating the expression:

I a [t - (sin 2t)/2]

Using the double-angle identity sin 2t 2 sin t cos t:

I a [t - sin t cos t]

Back Substitution

Now, substitute back x a sin2 t to express the integral in terms of x.

t sin?1 (√(x/a))

sin t √(x/a)

cos t √(1 - x/a) √((a-x)/a)

Therefore:

I a [sin?1 (√(x/a)) - √(x/a) * √((a-x)/a)] C

This simplifies to:

I a sin?1 (√(x/a)) - √(x(a-x)) C

Conclusion

The integral of ∫√[x/a - x] dx can be solved using a trigonometric substitution, leading to the expression:

I a sin?1 (√(x/a)) - √(x(a-x)) C

This solution requires a good understanding of trigonometric identities and integration techniques. For those interested in integral calculus, applying these techniques to solve integrals is a valuable skill.

Related Topics

Understanding and applying integration techniques in calculus is crucial. Here are some related topics to explore:

Integration by Parts

Integration by parts is another common technique used when dealing with products of functions.

Trigonometric Integrals

These integrals involve trigonometric functions and often require substitution or identities.

Improper Integrals

Improper integrals deal with unbounded intervals or unbounded functions, which often require careful analysis.