A) \[{{\tan }^{-1}}(\cos x)+c\]
B) \[-{{\tan }^{-1}}(\cos x)+c\]
C) \[{{\tan }^{-1}}(\sin x)+c\]
D) None of these
Correct Answer: B
Solution :
Let \[I=\int{\frac{\tan xdx}{\sec x+\cos x}}=\int{\frac{\frac{\sin x}{\cos x}}{\frac{1}{\cos x}+\cos x}}dx\] \[=\int{\frac{\sin x}{1+{{\cos }^{2}}x}}dx\] Let\[cos\text{ }x=t\] \[\Rightarrow \] \[-sin\text{ }x\text{ }dx=dt\] \[\therefore \] \[I=-\int{\frac{1}{1+{{t}^{2}}}}dt\] \[=-{{\tan }^{-1}}t+c\] \[=-{{\tan }^{-1}}(\cos x)+c\]
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