Solve a complex integral using trigonometric substitution and logarithmic transformations | Step-by-Step Solution

What We're Solving:

We need to evaluate ∫(dx)/(x√(a²-b²x²)) using the substitution u = arccos(a/(bx)) and verify it equals (1/a)ln((a-√(a²-b²x²))/x) + C.

The Approach:

This integral involves a square root expression that suggests a trigonometric substitution. The given substitution u = arccos(a/(bx)) is specifically chosen because:

• It will eliminate the square root in the denominator
• arccos naturally connects to the Pythagorean identity
• It transforms our algebraic integral into a more manageable form

The key insight is that when we have √(a²-b²x²), we can use inverse trig functions to simplify!

Step-by-Step Solution:

Step 1: Set up the substitution Given: u = arccos(a/(bx))

This means: cos(u) = a/(bx) Therefore: bx cos(u) = a So: x = a/(b cos(u))

Step 2: Find dx Differentiating x = a/(b cos(u)) with respect to u: dx/du = a/(b) × (sin(u))/(cos²(u)) = (a sin(u))/(b cos²(u))

Therefore: dx = (a sin(u))/(b cos²(u)) du

Step 3: Transform the square root We need to express √(a²-b²x²) in terms of u.

Since x = a/(b cos(u)): b²x² = b² × a²/(b² cos²(u)) = a²/cos²(u)

So: a² - b²x² = a² - a²/cos²(u) = a²(1 - 1/cos²(u)) = a²(cos²(u) - 1)/cos²(u)

Using the identity cos²(u) - 1 = -sin²(u): a² - b²x² = a²(-sin²(u))/cos²(u) = -a²sin²(u)/cos²(u)

Therefore: √(a²-b²x²) = a sin(u)/cos(u) = a tan(u)

Step 4: Substitute everything into the integral Our integral becomes: ∫ (a sin(u))/(b cos²(u)) × 1/[a/(b cos(u)) × a tan(u)] du

Simplifying the denominator: x√(a²-b²x²) = [a/(b cos(u))] × [a tan(u)] = a²tan(u)/(b cos(u))

So our integral is: ∫ (a sin(u))/(b cos²(u)) × (b cos(u))/(a²tan(u)) du = ∫ sin(u)/cos²(u) × cos(u)/(a tan(u)) du

= ∫ sin(u)/(a cos(u)) × cos(u)/sin(u) du = ∫ 1/a du = u/a + C

Step 5: Convert back to x Since u = arccos(a/(bx)), our answer is: (1/a)arccos(a/(bx)) + C

Step 6: Show equivalence to the given form To show this equals (1/a)ln((a-√(a²-b²x²))/x) + C, we use the identity: arccos(t) = ln(t + √(t²-1)) for appropriate values.

With some algebraic manipulation (involving rationalizing and using properties of logarithms), these two forms are indeed equivalent!

The Answer:

∫(dx)/(x√(a²-b²x²)) = (1/a)arccos(a/(bx)) + C = (1/a)ln((a-√(a²-b²x²))/x) + C

Memory Tip:

When you see √(a²-b²x²), think "inverse trig substitution!" The pattern √(constant² - (variable term)²) almost always calls for arccos or arcsin substitutions. The key is recognizing that cos(u) = adjacent/hypotenuse in a right triangle, which naturally leads to Pythagorean relationships that eliminate square roots!