Question

\(\int_0^a \frac{x}{\sqrt{a^2 - x^2}} \, dx \)

• A. \( \frac{a^2}{2} \)
• B. \( \frac{a}{2} \)
• C. \( \frac{a}{4} \)
• D. \( a \)

Hint:

When dealing with integrals of the form \( \int \frac{x}{\sqrt{a^2 - x^2}} \), use the trigonometric substitution \( x = a \sin \theta \).

Answer 1

The Correct Option is B

We are given: \[ I = \int_0^a \frac{x}{\sqrt{a^2 - x^2}} \, dx \] This is a standard integral. Let’s use the substitution \( x = a \sin \theta \), hence \( dx = a \cos \theta \, d\theta \), and \( \sqrt{a^2 - x^2} = a \cos \theta \). Now, substitute into the integral: \[ I = \int_0^{\frac{\pi}{2}} \frac{a \sin \theta}{a \cos \theta} a \cos \theta \, d\theta = a \int_0^{\frac{\pi}{2}} \sin \theta \, d\theta \] The integral of \( \sin \theta \) is \( -\cos \theta \), so: \[ I = a \left[ -\cos \theta \right]_0^{\frac{\pi}{2}} = a(0 - (-1)) = a \] Thus, the correct answer is \( \frac{a}{2} \).