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