A) \[{{\tan }^{-1}}(\sec x+\cos x)+C\]
B) \[{{\log }_{e}}|{{\tan }^{-1}}(\sec x+\cos x)|+C\]
C) \[\frac{1}{{{(\sec x+\cos x)}^{2}}}+C\]
D) None of the above
Correct Answer: B
Solution :
Let \[I=\int{\frac{{{\sin }^{3}}x}{({{\cos }^{4}}x+3{{\cos }^{2}}x+1){{\tan }^{-1}}(\sec x+\cos x)}dx}\]\[=\int{\frac{{{\sin }^{3}}x/{{\cos }^{2}}x}{({{\cos }^{2}}x+3+{{\sec }^{2}}x){{\tan }^{-1}}(\sec x+\cos x)}dx}\] \[=\int{\left[ \begin{align} & \frac{1}{1+{{(\sec x+\cos x)}^{2}}}\times \frac{\sin x(1-{{\cos }^{2}}x)}{{{\cos }^{2}}x} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\times \frac{1}{{{\tan }^{-1}}(\sec x+\cos x)} \\ \end{align} \right]}dx\] \[=\int{\left[ \begin{align} & \frac{1}{{{\tan }^{-1}}(\sec x+\cos x)}\times \frac{1}{1+{{(\sec x+\cos x)}^{2}}} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\times (\tan x\sec x-\sin x) \\ \end{align} \right]}dx\]\[=\int{\frac{1}{{{\tan }^{-1}}(\sec x+\cos x)}d\{{{\tan }^{-1}}(\sec x+\cos x)\}}\]. \[\therefore \] \[I={{\log }_{e}}|{{\tan }^{-1}}(\sec x+\cos x)|+C\]You need to login to perform this action.
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