A) \[2\,\,ta{{n}^{5}}\,\sqrt{x}+c\]
B) \[\frac{1}{5}\,ta{{n}^{5}}\,\sqrt{x}+c\]
C) \[\frac{2}{5} ta{{n}^{5}}\,\sqrt{x}+c\]
D) None of these
Correct Answer: C
Solution :
\[\int{\frac{1}{\sqrt{x}}{{\tan }^{4}}\sqrt{x}.{{\sec }^{2}}\sqrt{x}\,dx}\] Put tan \[\sqrt{x}=t\Rightarrow \frac{{{\sec }^{2}}\sqrt{x}}{2\sqrt{x}}dx=dt\], then it reduces to to \[2\int{{{t}^{4}}dt=\frac{2}{5}{{(tan\sqrt{x})}^{5}}+c=\frac{2}{5}{{\tan }^{5}}\,\sqrt{x}+c}\]
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