Integral Calculus Question 421
Question: $ \int_{{}}^{{}}{{{\sec }^{2/3}}x,cose{c^{4/3}}x\ dx=} $
Options:
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 $
Show Answer
Answer:
Correct Answer: B
Solution:
$ \int_{{}}^{{}}{{{\sec }^{2/3}}x,cose{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 $ .
BETA
