A) \[\frac{\pi }{4}\]
B) 0
C) 1
D) \[-\frac{\pi }{4}\]
Correct Answer: A
Solution :
Consider \[\int\limits_{0}^{{{\sin }^{2}}x}{{{\sin }^{-1}}(\sqrt{t})dt+\int\limits_{0}^{{{\cos }^{2}}x}{{{\cos }^{-1}}(\sqrt{t})dt}}\] Let \[I=f(x)\] after integrating and putting the limits. \[f'(x)={{\sin }^{-1}}\sqrt{{{\sin }^{2}}}x(2\sin x\cos x)-0\] \[+{{\cos }^{-1}}\sqrt{{{\cos }^{2}}x}(-2\cos x\sin x)-0\] \[\therefore \]\[f(x)=0\Rightarrow f(x)=C\](constant) Now, we find/(x) at \[x=\frac{\pi }{4}\] \[\therefore \]\[I=\int\limits_{0}^{1/2}{{{\sin }^{-1}}}\sqrt{t}dt+\int\limits_{0}^{1/2}{{{\cos }^{-1}}}\sqrt{t}dt\] \[=\int\limits_{0}^{1/2}{({{\sin }^{-1}}}\sqrt{t}+{{\cos }^{-1}}\sqrt{t})dt\] \[=\int\limits_{0}^{1/2}{\frac{\pi }{2}dt}=\frac{\pi }{4}=C\] \[\therefore \]\[f(x)=\frac{\pi }{4}\] \[\therefore \]Required integration = ?
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