_ ^ ^ 2/3 x , cose c ^ 4/3 x ;dx = - Vidyadip

MCQ

$\int_{}^{} {{{\sec }^{2/3}}x\,{\rm{cose}}{{\rm{c}}^{4/3}}x\;dx = } $

A
$ - 3{(\tan x)^{1/3}} + c$
✓
$ - 3{(\tan x)^{ - 1/3}} + c$
C
$3{(\tan x)^{ - 1/3}} + c$
D
${(\tan x)^{ - 1/3}} + c$

✓

Answer

Correct option: B.

$ - 3{(\tan x)^{ - 1/3}} + c$

b
(b)$\int_{}^{} {{{\sec }^{2/3}}x\,{\rm{cose}}{{\rm{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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