A) \[\frac{3}{2}{{\sin }^{-1}}x-\frac{1}{2}x\sqrt{1-{{x}^{2}}}+c\]
B) \[\frac{3}{2}{{\sin }^{-1}}x+\frac{1}{2}x\sqrt{1-{{x}^{2}}}+c\]
C) \[\frac{3}{2}{{\cos }^{-1}}x-\frac{1}{2}x\sqrt{1-{{x}^{2}}}+c\]
D) \[\frac{3}{2}{{\cos }^{-1}}x+\frac{1}{2}x\sqrt{1-{{x}^{2}}}+c\]
Correct Answer: A
Solution :
Put \[x=\sin \theta \Rightarrow dx=\cos \theta \,d\theta ,\] then it reduces to \[\int_{{}}^{{}}{(1+{{\sin }^{2}}\theta )\,d\theta }=\theta +\frac{1}{2}\int_{{}}^{{}}{(1-\cos 2\theta )\,d\theta }\] \[=\frac{3\theta }{2}-\frac{1}{2}\sin \theta \sqrt{1-{{\sin }^{2}}\theta }+c=\frac{3}{2}{{\sin }^{-1}}x-\frac{1}{2}x\sqrt{1-{{x}^{2}}}+c\].
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