On solving , Anil's answer was and Jaspreet's answer was . Whose answer was correct?
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
Answer
AIWritten by AI - it can contain mistakes.
Correct option: c
(c) Both of them were correct.
We know that $\frac{d}{dx}(\cos^{-1}x) = -\frac{1}{\sqrt{1-x^2}}$ and $\frac{d}{dx}(\sin^{-1}x) = \frac{1}{\sqrt{1-x^2}}$.
Thus, $\int -\frac{3}{\sqrt{1-x^2}} dx = 3\cos^{-1}x + c_1$, and also $-3\int \frac{dx}{\sqrt{1-x^2}} = -3\sin^{-1}x + c_2$.
Since $\cos^{-1}x + \sin^{-1}x = \frac{\pi}{2}$, $3\cos^{-1}x = 3(\frac{\pi}{2} - \sin^{-1}x) = -3\sin^{-1}x + \frac{3\pi}{2}$.
The two results differ only by an arbitrary constant, so both are correct.
From ISC 2027 Specimen Mathematics Paper 1, question 1(x).