Indefinite Integration Question - None

Question

The integral $\int x\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)dx\;(x>0)$ is equal to

Accepted Answer

Step 1: Use the identity cos⁻¹((1-x²)/(1+x²)) = 2tan⁻¹(x) for x > 0

This comes from cos(2θ) = (1-tan²θ)/(1+tan²θ), so if tan⁻¹(x) = θ, then cos⁻¹((1-x²)/(1+x²)) = 2θ = 2tan⁻¹(x)

Step 2: Rewrite the integral as ∫ x · 2tan⁻¹(x) dx = 2∫ x·tan⁻¹(x) dx

Step 3: Apply integration by parts with u = tan⁻¹(x), dv = x dx

Then du = 1/(1+x²) dx, v = x²/2

Step 4: ∫ x·tan⁻¹(x) dx = (x²/2)tan⁻¹(x) - ∫ (x²/2) · 1/(1+x²) dx

Step 5: Simplify the second integral: ∫ x²/(2(1+x²)) dx = (1/2)∫ (1+x²-1)/(1+x²) dx = (1/2)[x - tan⁻¹(x)] + C

Step 6: Combine results: 2[(x²/2)tan⁻¹(x) - (1/2)(x - tan⁻¹(x))] + C

= (x²+1)tan⁻¹(x) - x + C

∴ Answer: (x²+1)tan⁻¹(x) - x + C

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