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

Explanation for correct answer:Find the relation:Taking I_2 first,I_2 =∫sin^ 1√((1–x^2)) dx0ex0ex =∫cos^ 1xdxI_1=∫sin^ 1xdx (Given)I_1+I_2=∫sin^ 1xd ...

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Byju's Answer

Standard XII

Mathematics

Integration by Parts

If I1=∫sin-1 ...

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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$I_{1} + I_{2} = \frac{π}{2} x + C$

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$I_{1} + I_{2} = \frac{π}{2} + C$

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None of these

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Solution

The correct option is D
$I_{1} + I_{2} = \frac{π}{2} + C$

Explanation for correct answer:
Find the relation:
Taking $I_{2}$ first,
$I_{2} = \int \sin^{- 1} \sqrt{(1 - x^{2})} d x = \int \cos^{- 1} x d x$
$I_{1} = \int \sin^{- 1} x d x$ (Given)
$I_{1} + I_{2} = \int \sin^{- 1} x d x + \int \cos^{- 1} x d x = \int (\sin^{- 1} x + \cos^{- 1} x) d x = \frac{π}{2} \int d x [∵ s i n^{- 1} + c o s^{- 1} x = \frac{π}{2}] = \frac{π}{2} x + C$
Hence, the correct option is C.

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If I1=∫sin-1xdxand I2=∫sin-1√(1-x2)dx, then

The correct option is D I1+I2=π2+CExplanation for correct answer:Find the relation:Taking I2 first,I2=∫sin-1(1–x2)dx=∫cos-1xdxI1=∫sin-1xdx (Given)I1+I2=∫sin-1xdx+∫cos-1xdx=∫(sin-1x+cos-1x)dx=π2∫dx[∵sin-1+cos-1x=π2]=π2x+CHence, the correct option is C.

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