If I1 - -sin-1 x dx and I2 - -sin-1-1 - x2 dx- then

The correct option is C I_1 + I_2 = π2x Given, I_1 = ∫sin^ 1 x dx and I_2 = ∫sin^ 1√(1 x^2) dx ⇒ I_2 = ∫cos^ 1 x dx Now, I_1 + I_2 = ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Equations

If I1 = ∫si...

Question

If $I_{1} = \int {sin}^{- 1} x d x$ and $I_{2} = \int {sin}^{- 1} \sqrt{1 - x^{2}} d x$ , then

I_{1} = I_{2}

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I_{2} = \frac{π}{2} I_{1}

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I_{1} + I_{2} = \frac{π}{2} x

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I_{1} + I_{2} = \frac{π}{2}

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Solution

The correct option is C $I_{1} + I_{2} = \frac{π}{2} x$
Given, $I_{1} = \int {sin}^{- 1} x d x$
and $I_{2} = \int {sin}^{- 1} \sqrt{1 - x^{2}} d x$
$\Rightarrow I_{2} = \int {cos}^{- 1} x d x$
Now, $I_{1} + I_{2} = \int ({sin}^{- 1} x + {cos}^{- 2} x) d x$
$= \int \frac{π}{2} d x = \frac{π}{2} x$
$∴ I_{1} + I_{2} = \frac{π}{2} x$

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If I1=∫sin−1xdx and I2=∫sin−1√1−x2dx, then

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If $I_{1} = \int {sin}^{- 1} x d x$ and $I_{2} = \int {sin}^{- 1} \sqrt{1 - x^{2}} d x$ , then