A) \[\sqrt{2}-1\]
B) \[1-\sqrt{2}\]
C) \[\sqrt{2}+1\]
D) None of these
Correct Answer: A
Solution :
Let \[\int_{\pi /4}^{\pi /2}{\cos \theta \frac{1}{{{\sin }^{2}}\theta }d\theta }\] Put \[t=\sin \theta \Rightarrow dt=\cos \theta \,\,d\theta ,\] then we have \[\int_{1/\sqrt{2}}^{1}{\frac{1}{{{t}^{2}}}dt}=\left[ \frac{-1}{t} \right]_{1/\sqrt{2}}^{1}=\sqrt{2}-1\].
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