I_1 = ^ - 1 x , ,dx and I_2 = ^ - 1 1 - x^2 dxthen

MCQ

${I_1} = \int {{{\sin }^{ - 1}}x\,\,dx} $ and ${I_2} = \int {{{\sin }^{ - 1}}\sqrt {1 - {x^2}} } dx$then

A
${I_1} = {I_2}$
B
${I_2} = \frac{\pi }{2}I_1$
✓
${I_1} + {I_2} = \frac{\pi }{2}x$
D
${I_1} + {I_2} = \pi /2$

✓

Answer

Correct option: C.

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

c
(c) ${I_1} = \int {{{\sin }^{ - 1}}xdx} $
Let ${\sin ^{ - 1}}x = \theta $==> $x = \sin \theta $ ==> $dx = \cos \theta \,d\theta $
${I_1} = \int {\theta \cos \theta d\theta } $$ = \theta \sin \theta - \int {\sin \theta d\theta } $$ = \theta \sin \theta + \cos \theta $
$ = x{\sin ^{ - 1}}x + \sqrt {1 - {x^2}} $
${I_2} = \int {{{\sin }^{ - 1}}\sqrt {1 - {x^2}} } dx$$ = \int {{{\cos }^{ - 1}}xdx} $
Let $\cos \phi = x,$ Hence $ - \sin \phi \,d\phi = dx$
${I_2} = - \int {\phi hi \sin \phi d\phi } $$ = \phi \cos \phi + \int { - \cos \phi d\phi } $
$ = \phi \cos \phi - \sin \phi $$ = x{\cos ^{ - 1}}x - \sqrt {1 - {x^2}} $
${I_1} + {I_2} = x({\cos ^{ - 1}}x + {\sin ^{ - 1}}x) = \frac{\pi }{2}x$.

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