A) \[2{{\tan }^{5}}\sqrt{x}+c\]
B) \[\frac{1}{5}{{\tan }^{5}}\sqrt{x}+c\]
C) \[\frac{2}{5}{{\tan }^{5}}\sqrt{x}+c\]
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 \[2\int_{{}}^{{}}{{{t}^{4}}dt}=\frac{2}{5}{{(\tan \sqrt{x})}^{5}}+c=\frac{2}{5}{{\tan }^{5}}\sqrt{x}+c\].
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