Q91 / 150 · MHT CET 2026 (18 April Shift 1) · Mathematics
If $0 \leq x \leq 1$, $I_1 = \int\sin^{-1}\sqrt{1-x^2}\,dx$ and $I_2 = \int\sin^{-1}x\,dx$, then which of the following is true?
- A$I_1 = I_2$
- B$I_1 = \dfrac{\pi}{2}I_2$
- Correct answerC$I_1 + I_2 = \dfrac{\pi}{2}x$
- D$I_1 + I_2 = \dfrac{\pi}{2}$
Solution
For $0 \leq x \leq 1$, the inverse trigonometric identity is:
$\sin^{-1}\sqrt{1-x^2} = \cos^{-1}x$
Substituting this into $I_1$:
$I_1 = \int \cos^{-1}x \,dx$
Given:
$I_2 = \int \sin^{-1}x \,dx$
Adding $I_1$ and $I_2$:
$I_1 + I_2 = \int \cos^{-1}x \,dx + \int \sin^{-1}x \,dx$
$I_1 + I_2 = \int (\cos^{-1}x + \sin^{-1}x) \,dx$
Using the identity $\sin^{-1}x + \cos^{-1}x = \dfrac{\pi}{2}$:
$I_1 + I_2 = \int \dfrac{\pi}{2} \,dx$
$I_1 + I_2 = \dfrac{\pi}{2}x$
Answer: $I_1 + I_2 = \dfrac{\pi}{2}x$
$\sin^{-1}\sqrt{1-x^2} = \cos^{-1}x$
Substituting this into $I_1$:
$I_1 = \int \cos^{-1}x \,dx$
Given:
$I_2 = \int \sin^{-1}x \,dx$
Adding $I_1$ and $I_2$:
$I_1 + I_2 = \int \cos^{-1}x \,dx + \int \sin^{-1}x \,dx$
$I_1 + I_2 = \int (\cos^{-1}x + \sin^{-1}x) \,dx$
Using the identity $\sin^{-1}x + \cos^{-1}x = \dfrac{\pi}{2}$:
$I_1 + I_2 = \int \dfrac{\pi}{2} \,dx$
$I_1 + I_2 = \dfrac{\pi}{2}x$
Answer: $I_1 + I_2 = \dfrac{\pi}{2}x$
Hence, the correct answer is option: 'C'.