A) \[\frac{1}{2(b-a)}\log (a{{\cos }^{2}}x+b{{\sin }^{2}}x)+c\]
B) \[\frac{1}{b-a}\log (a{{\cos }^{2}}x+b{{\sin }^{2}}x)+c\]
C) \[\frac{1}{2}\log (a{{\cos }^{2}}x+b{{\sin }^{2}}x)+c\]
D) None of these
Correct Answer: A
Solution :
Put \[a{{\cos }^{2}}x+b{{\sin }^{2}}x=t\]\[\Rightarrow 2(b-a)\sin x\cos x=dt,\] then \[\int_{{}}^{{}}{\frac{\sin x\cos x\,dx}{a{{\cos }^{2}}x+b{{\sin }^{2}}x}=\frac{1}{2(b-a)}\int_{{}}^{{}}{\frac{1}{t}\,dt}}\] \[=\frac{1}{2(b-a)}\log (a{{\cos }^{2}}x+b{{\sin }^{2}}x)+c\].
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