Question

The integral ∫xcos^-1((1 - x²)/(1 + x²))dx(x > 0)  is equal to

(A) −x + (1 + x²) tan⁻¹ x + c 

(B) x − (1 + x²) cot⁻¹ x + c 

(C) −x + (1 + x²) cot⁻¹ x + c 

(D) x − (1 + x²) tan⁻¹ x + c

Answer 1

Answer is (A) −x + (1 + x²)tan⁻¹x + c

Again put tanθ = t.

Then sec²θdθ = dt.

Therefore

= t² tan⁻¹t − [t − tan⁻¹t] = t²tan⁻¹t − t + tan⁻¹t = (t² + 1) tan⁻¹t − t = {tan²θ + 1} tan⁻¹ (tanθ ) − tanθ = θ{tan²θ + 1} − tanθ 

Therefore, I = (1 + x²) tan⁻¹x − x + c.