Question

Prove that \( \displaystyle \int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) + C \).

Hint:

For integrals of the form \( \sqrt{a^2 - x^2} \), always try substitution \( x = a \sin \theta \).

Answer 1

To Prove:
∫ √(a² − x²) dx = (x/2)√(a² − x²) + (a²/2) sin⁻¹(x/a) + C

Step 1: Use Trigonometric Substitution
Let x = a sin θ

Then,

dx = a cos θ dθ

Also,

√(a² − x²) = √(a² − a² sin²θ)

= √(a²(1 − sin²θ))

= √(a² cos²θ)

= a cos θ

Step 2: Substitute in the Integral

∫ √(a² − x²) dx

= ∫ (a cos θ)(a cos θ dθ)

= a² ∫ cos²θ dθ

Step 3: Use Identity

cos²θ = (1 + cos2θ)/2

Therefore,

a² ∫ cos²θ dθ

= a² ∫ (1 + cos2θ)/2 dθ

= (a²/2) ∫ (1 + cos2θ) dθ

= (a²/2) [ θ + (1/2) sin2θ ] + C

Step 4: Simplify

sin2θ = 2 sinθ cosθ

So,

(a²/2) [ θ + sinθ cosθ ] + C

Step 5: Convert Back to x

Since x = a sinθ,

sinθ = x/a

cosθ = √(a² − x²)/a

θ = sin⁻¹(x/a)

Therefore,

sinθ cosθ = (x/a)(√(a² − x²)/a)

= x√(a² − x²)/a²

Substitute back:

(a²/2)[ sin⁻¹(x/a) + x√(a² − x²)/a² ] + C

= (a²/2) sin⁻¹(x/a)

+ (1/2) x√(a² − x²) + C

Final Result:
∫ √(a² − x²) dx

= (x/2)√(a² − x²) + (a²/2) sin⁻¹(x/a) + C

Hence proved.