A) \[{{\cos }^{-1}}\left( \frac{19}{35} \right)\]
B) \[{{\cos }^{-1}}\left( \frac{17}{31} \right)\]
C) \[100\pi c{{m}^{3}}/\min \]
D) \[{{T}^{2}}\propto {{R}^{3}}\Rightarrow \frac{{{T}_{2}}}{{{T}_{1}}}={{\left( \frac{{{R}_{2}}}{{{R}_{1}}} \right)}^{3/2}}={{\left( \frac{3R}{R} \right)}^{3/2}}\]
Correct Answer: A
Solution :
Let \[I=\int{\frac{dx}{({{x}^{2}}+1)({{x}^{2}}+4)}}\] \[=\frac{1}{3}\int{\left( \frac{1}{{{x}^{2}}+1}-\frac{1}{{{x}^{2}}+4} \right)}dx\] \[=\frac{1}{3}{{\tan }^{-1}}x-\frac{1}{6}{{\tan }^{-1}}\frac{1}{2}\] \[=k{{\tan }^{-1}}x-1{{\tan }^{-1}}\frac{x}{2}\] \[\therefore \] \[k=\frac{1}{3}and1=-\frac{1}{6}\]
You need to login to perform this action.
You will be redirected in
3 sec
