A) \[\log \sin 2x+c\]
B) \[\log (1+{{\sin }^{2}}x)+c\]
C) \[\frac{1}{2}\log (1+{{\sin }^{2}}x)+c\]
D) \[{{\tan }^{-1}}(\sin x)+c\]
Correct Answer: B
Solution :
Put \[(1+{{\sin }^{2}}x)=t\Rightarrow \sin 2x\,dx=dt\] Hence \[\int_{{}}^{{}}{\frac{\sin 2x}{1+{{\sin }^{2}}x}dx}=\int_{{}}^{{}}{\frac{1}{t}dt=\log (1+{{\sin }^{2}}x)+c.}\]
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