A) \[\frac{\pi }{2}\]
B) \[1\]
C) \[\frac{\pi }{4}\]
D) None of these
Correct Answer: B
Solution :
Let\[f(x)=\int_{0}^{{{\sin }^{2}}x}{{{\sin }^{-1}}\sqrt{t}dt}+\int_{0}^{{{\cos }^{2}}x}{{{\cos }^{-1}}}\sqrt{t}dt\] \[\Rightarrow \] \[f'(x)={{\sin }^{-1}}(\sqrt{{{\sin }^{2}}x})\cdot 2\sin x\cdot \cos x\] \[+{{\cos }^{-1}}(\sqrt{{{\cos }^{2}}t})\cdot (2-\cos x\cdot \sin x)\] \[=\sin 2x\cdot \{{{\sin }^{-1}}(\sin x)\}+(-\sin 2x)\{{{\cos }^{-1}}(\cos x)\}\] \[=\sin 2x(x-x)=0\] \[\therefore \] \[f'(x)=0\] \[\Rightarrow \] \[f(x)=\]constant Find \[f(x)\] for any value of\[x,\] \[\int\limits_{0}^{1/2}{\frac{\pi }{2}dx}=\frac{\pi }{2}\cdot \frac{1}{2}=\frac{\pi }{4}\]
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