A) \[\frac{1}{3}{{\tan }^{2}}x+c\]
B) \[\frac{1}{2}{{\tan }^{2}}x+c\]
C) \[\frac{1}{3}{{\tan }^{3}}x+c\]
D) \[3\sin 2x-4\cos 4x+c\]
E) \[{{\cos }^{3}}x+c\]
Correct Answer: C
Solution :
Let\[I=\int{\frac{{{\sin }^{2}}x}{{{\cos }^{4}}x}}dx\] \[=\int{{{\tan }^{2}}x{{\sec }^{2}}x\,dx}\] Let \[tan\text{ }x=t\] \[\Rightarrow \] \[se{{c}^{2}}x\text{ }dx=dt\] \[\therefore \]\[I=\int{{{t}^{2}}}\,dt=\frac{{{t}^{3}}}{3}+c=\frac{{{\tan }^{3}}x}{3}+c\]
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