If {I_1} = int {{{sin }^{ - 1}}x,dx} and {I_ | vedclass

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(C) Given ${I_1} = \int {{{\sin }^{ - 1}}x\,dx} $ and ${I_2} = \int {{{\sin }^{ - 1}}\sqrt {1 - {x^2}} } dx$. Using the identity ${\sin ^{ - 1}}\sqrt

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

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(C) Given ${I_1} = \int {{{\sin }^{ - 1}}x\,dx} $ and ${I_2} = \int {{{\sin }^{ - 1}}\sqrt {1 - {x^2}} } dx$. Using the identity ${\sin ^{ - 1}}\sqrt {1 - {x^2}} = {\cos ^{ - 1}}x$,we have ${I_2} = \int {{{\cos }^{ - 1}}x\,dx} $. Now,consider the sum ${I_1} + {I_2} = \int {{{\sin }^{ - 1}}x\,dx} + \int {{{\cos }^{ - 1}}x\,dx} $. Since ${\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \frac{\pi }{2}$,we get ${I_1} + {I_2} = \int {(\sin ^{ - 1}x + \cos ^{ - 1}x)\,dx} $. ${I_1} + {I_2} = \int {\frac{\pi }{2}\,dx} = \frac{\pi }{2}x + C$. Assuming the constant of integration $C = 0$ for the relation,we get ${I_1} + {I_2} = \frac{\pi }{2}x$.

List-$I$	List-$II$
$1. \int \frac{\sin^2 x}{\cos^4 x} dx$	$A. \frac{\tan^2 x}{2} + \ln\|\cos x\| + c$
$2. \int \frac{\sin^4 x}{\cos^2 x} dx$	$B. \cos x + \sec x + c$
$3. \int \frac{\sin^3 x}{\cos^2 x} dx$	$C. \frac{\tan^3 x}{3} + c$
$4. \int \frac{\sin^3 x}{\cos^3 x} dx$	$D. \tan x + \frac{\sin 2x}{4} - \frac{3x}{2} + c$
	$E. \cos x - \sec x + c$

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

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