If 0 ≤ x ≤ 1, I₁= ^{-1} {1-x²} d x and I₂= ^{-1} x d x, then which of the following is true?

Correct Answer is : I1+I2=π2xI₁+I₂=π/2 xI1+I2=2πx

Correct Answer is : I1+I2=π2xI₁+I₂=π/2 xI1+I2=2πx

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MHTCET 18 Apr 2026 Shift 1 Paper

Question : 117 of 150

Marks: +1, -0

0 \leq x \leq 1, I_{1}=\int \sin ^{-1} \sqrt{1-x^{2}} d x

I_{2}=\int \sin ^{-1} x d x

, then which of the following is true?

Solution:

Concept:

For

0 \leq x \leq 1

, substitute

\theta = \sin^{-1} x

to simplify

\sin^{-1}\sqrt{1-x^2}

using the identity

\sin^{-1}(\cos \theta) = \frac{\pi}{2} - \theta

Explanation:

Let

\theta = \sin^{-1} x

Since

0 \leq x \leq 1

, we have

0 \leq \theta \leq \frac{\pi}{2}

Then

x = \sin \theta

, so

\sqrt{1-x^2} = \sqrt{1-\sin^2\theta} = \cos \theta.

Therefore,

\sin^{-1}\sqrt{1-x^2} = \sin^{-1}(\cos \theta) = \frac{\pi}{2} - \theta = \frac{\pi}{2} - \sin^{-1}x.

Substitute this into

I_1

I_1 = \int \left(\frac{\pi}{2} - \sin^{-1}x\right)\,dx.

Add

I_2 = \int \sin^{-1}x\,dx

I_1 + I_2 = \int \frac{\pi}{2}\,dx = \frac{\pi}{2}x + C.

Ignoring the arbitrary constant

C

, as intended in the options, we get

I_1 + I_2 = \frac{\pi}{2}x

Answer:

Option C is correct:

I_1 + I_2 = \frac{\pi}{2}x

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