A) \[-\frac{1}{2}\log \,(1+{{\cos }^{2}}x)+c\]
B) \[2\,\,\log \,(1+{{\cos }^{2}}x)+c\]
C) \[\frac{1}{2}\,\,\log \,(1+{{\cos }^{2}}2x)+c\]
D) \[c-\,\log \,\,(1+{{\cos }^{2}}x)\]
Correct Answer: D
Solution :
Let \[I=\int{\frac{2\sin x\cos x}{1+{{\cos }^{2}}x}d\,x}\] Put \[1+{{\cos }^{2}}x=t\] \[\therefore \] \[-2\cos x\sin x\,dx=dt\] \[\therefore \] \[-I\int{\frac{-dt}{t}=-\log \,\,t+c}\] \[=-\log \,\,(1+{{\cos }^{2}}x)+c\]
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