A) \[\frac{2}{3}\]
B) \[-\frac{2}{3}\]
C) \[\frac{3}{2}\]
D) \[-\frac{3}{2}\]
Correct Answer: A
Solution :
Idea Understand the basic concept of modulus function. Here \[\because \] we know that \[-1\le \sin x\le \]1 for \[-\frac{\pi }{2}<x<\frac{\pi }{2}\]but where \[-\frac{\pi }{2}<x<0.\] Then \[-1\le \sin x\le 0.\] Hence, we know that \[{{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1\] We have given that \[\int_{-\frac{\pi }{2}}^{0}{\sqrt{\cos x(1-{{\cos }^{2}}x)}dx}\] \[\int_{-\pi /2}^{0}{\sqrt{\cos x}|\sin x|dx}\] when \[-\frac{\pi }{2}<x<0\]then\[-1\le \sin x\le 0\] \[\therefore \]\[-\int_{-\pi /2}^{0}{\sqrt{\cos x}\sin xdx}\] Put \[\cos x=t,-\sin xdx=dt\] \[\int_{0}^{1}{{{t}^{1/2}}}dt\] \[\left( \frac{{{t}^{3/2}}}{3/2} \right)_{0}^{1}=+\frac{2}{3}\] TEST Edge Let JEE Main properties of definite integrals based questions are asked to solve such types of questions students are advised that learn properties of definite integral and understand the concept of integration.
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