A) \[\frac{{{a}^{2}}}{4}\]
B) \[\frac{\pi }{2}\]
C) \[\frac{\pi }{4}\]
D) \[\pi \]
E) \[\frac{{{\pi }^{2}}}{4}\]
Correct Answer: C
Solution :
Let \[I=\int_{0}^{a}{\frac{dx}{x+\sqrt{{{a}^{2}}-{{x}^{2}}}}}\] Let \[x=a\text{ }sin\theta \] \[\Rightarrow \] \[dx=a\text{ }cos\theta \text{ }d\theta \] \[\therefore \] \[I=\int_{0}^{\pi /2}{\frac{a\cos \theta d\theta }{a\sin \theta +a\cos \theta }}\] \[=\int_{0}^{\pi /2}{\frac{\cos \theta }{\sin \theta +\cos \theta }}d\theta \] ?. (i) \[\Rightarrow \] \[I=\int_{0}^{\pi /2}{\frac{\cos (\pi /2-\theta )}{\sin (\pi /2-\theta )+\cos (\pi /2-\theta )}}d\theta \] \[\therefore \] \[I=\int_{0}^{\pi /2}{\frac{\cos \theta }{\sin \theta +\cos \theta }}d\theta \] ?.(ii) On adding Eqs. (i) and (ii), we get \[2I=\int_{0}^{\pi /2}{1\,}d\theta =[\theta ]_{0}^{\pi /2}\] \[\Rightarrow \] \[I=\frac{\pi }{4}\]
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