A) \[\frac{1}{15}{{\tan }^{-1}}\,\left( \frac{2\,\,\tan \,\,x}{\sqrt{5}} \right)+c\]
B) \[\frac{1}{\sqrt{5}}{{\tan }^{-1}}\,\left( \frac{\tan \,\,x}{\sqrt{5}} \right)+c\]
C) \[\frac{1}{2\sqrt{5}}\,{{\tan }^{-1}}\,\left( \frac{2\,\,\tan \,\,x}{\sqrt{5}} \right)+c\]
D) None of these
Correct Answer: C
Solution :
\[\int{\frac{dx}{4{{\sin }^{2}}x+5{{\cos }^{2}}x}}=\int{\frac{\,{{\sec }^{2}}xdx}{4\,{{\tan }^{2}}x+5}}=\frac{1}{4}\int{\frac{{{\sec }^{2}}xdx}{{{\tan }^{2}}x+\frac{5}{4}}}\]Put \[\tan x =t\,\,\,\Rightarrow \,\,se{{c}^{2}}\,x\,dx=\,\,dt\], then it reduces to \[\frac{1}{4}\int{\frac{dt}{{{t}^{2}}+{{\left( \frac{\sqrt{5}}{2} \right)}^{2}}}}=\frac{2}{4\sqrt{5}}{{\tan }^{-1}}\left( \frac{2t}{\sqrt{5}} \right)+c\] \[=\,\,\frac{1}{2\sqrt{5}}{{\tan }^{-1}}\,\left( \frac{2\,\tan \,x}{\sqrt{5}} \right)+c\]
You need to login to perform this action.
You will be redirected in
3 sec
