A) \[\sqrt{\cos \,\,x}+C\]
B) \[2\sqrt{\cot \,\,x}+C\]
C) \[\sqrt{\tan \,\,x}+C\]
D) \[2\sqrt{\tan \,\,x}+C\]
Correct Answer: D
Solution :
Let \[I=\int{\frac{\sqrt{\tan \,x}}{\sin \,\,x\cos \,\,x}}\,dx\] Divided by \[{{\cos }^{2}}x\,\,to\,\,{{N}^{r}}\,\,and\,\,{{D}^{r}}\] \[=\,\,\,\int{\frac{\sqrt{\tan \,\,x}/{{\cos }^{2}}\,x}{\sin x\cos x/{{\cos }^{2}}x}}\,dx\] \[=\,\int{\frac{\sqrt{\tan \,\,x}/{{\sec }^{2}}\,x}{\tan \,\,x}}\,dx=\int{\,\frac{{{\sec }^{2}}x}{\sqrt{\tan \,x}}}\,dx\] Put \[\tan \,\,x=t\], \[\Rightarrow \,\,{{\sec }^{2}}x\,dx=dt\], \[\therefore \,\,\,I=\int{\frac{1}{\sqrt{t}}}\,dt=2\sqrt{t}+C\] \[\Rightarrow \,\,\,\,I=2\sqrt{\tan x}+C\]
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