On solving ∫ -3/(√(1-x²)) dx, Anil's answer was ∫ -3/(√(1-x²)) dx = 3 cos⁻¹ x + c and Jaspreet's answer was ∫ -3/(√(1-x²)) dx = -3 ∫ (dx)/(√(1-x²)) = -3…

Answer: (c) Both of them were correct. cos⁻¹x = π/2 - sin⁻¹x, so 3cos⁻¹x + c = -3sin⁻¹x + ((3π)/2 + c), which differs from Jaspreet's answer only by a constant. Correct: c

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Mathematics20271 markMCQIntegrals

Choose the correct option.

On solving ∫ -3/(√(1-x²)) dx, Anil's answer was ∫ -3/(√(1-x²)) dx = 3 cos⁻¹ x + c and Jaspreet's…

On solving $\int -\frac{3}{\sqrt{1-x^2}} dx$, Anil's answer was $\int -\frac{3}{\sqrt{1-x^2}} dx = 3 \cos^{-1} x + c$ and Jaspreet's answer was $\int -\frac{3}{\sqrt{1-x^2}} dx = -3 \int \frac{dx}{\sqrt{1-x^2}} = -3 \sin^{-1} x + c$. Whose answer was correct?

aOnly Anil's answer was correct.
bOnly Jaspreet's answer was correct.
cBoth of them were correct.
dNeither of them was correct.

Answer

Official answer key

Correct option: c

Answer: (c) Both of them were correct. $\cos^{-1}x = \frac{\pi}{2} - \sin^{-1}x$, so $3\cos^{-1}x + c = -3\sin^{-1}x + \left(\frac{3\pi}{2} + c\right)$, which differs from Jaspreet's answer only by a constant.