A) \[{{\tan }^{4}}x+C\]
B) \[\tan 4x+C\]
C) \[{{\tan }^{4}}2x+x+C\]
D) \[\frac{1}{8}{{\tan }^{4}}2x+C\]
Correct Answer: D
Solution :
\[I=\int{\frac{{{\sin }^{3}}2x}{{{\cos }^{5}}2x}dx}\] Þ \[I=\int{\frac{{{\sin }^{3}}2x}{{{\cos }^{3}}2x}.\frac{1}{{{\cos }^{2}}2x}dx=\int{{{\tan }^{3}}2x.{{\sec }^{2}}2x\,dx.}}\] Putting tan \[2x=t\] and \[2{{\sec }^{2}}2x\,dx=dt\], we get \[I=\int{{{t}^{3}}\frac{dt}{2}=\frac{1}{2}.\frac{{{t}^{4}}}{4}+C=\frac{1}{8}({{\tan }^{4}}2x)+C.}\]
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