A) \[-3{{(\tan x)}^{1/3}}+c\]
B) \[-3{{(\tan x)}^{-1/3}}+c\]
C) \[3{{(\tan x)}^{-1/3}}+c\]
D) \[{{(\tan x)}^{-1/3}}+c\]
Correct Answer: B
Solution :
\[\int_{{}}^{{}}{{{\sec }^{2/3}}x\,\text{cose}{{\text{c}}^{4/3}}x\,dx}=\int_{{}}^{{}}{\frac{dx}{{{\sin }^{4/3}}x{{\cos }^{2/3}}x}}\] Multiplying \[{{N}^{r}}\] and \[{{D}^{r}}\] by \[{{\cos }^{2}}x,\] we get {Putting \[\tan x=t\Rightarrow {{\sec }^{2}}x\,dx=dt\}\] \[=\int_{{}}^{{}}{\frac{{{\sec }^{2}}x\,dx}{{{\tan }^{4/3}}x}}=\int_{{}}^{{}}{\frac{dt}{{{t}^{4/3}}}}=\frac{{{t}^{-1/3}}}{(-1/3)}+c=-3{{(\tan x)}^{-1/3}}+c\].
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