Question

Determine the integration $\int {{{\operatorname{Sec} }^{4/3}}} x.\cos e{c^{8/3}}xdx$ is equal to
A. $ - \dfrac{3}{5}{\tan ^{ - 8/3}}x + 3{\tan ^{ - 2/3}}x + C$
B. $ - \dfrac{3}{5}{\tan ^{ - 5/3}}x + 3{\tan ^{1/3}}x + C$
C. $\dfrac{3}{5}{\tan ^{ - 8/3}}x + 3{\tan ^{ - 2/3}}x + C$

Answer 1

Hint: First of all we have to determine the integration $\int {{{\operatorname{Sec} }^{4/3}}} x.\cos e{c^{8/3}}xdx$. First of all we have to change the $\sec x$ and $\cos ecx$ into the $\cos x$ and $\sin x$ respectively. With the help of the formula below:
Formula used:
$\sec x$= $\dfrac{1}{{\operatorname{Cos} x}}$and $\cos ecx$= $\dfrac{1}{{\sin x}}$
Now, we have to multiply by ${\cos ^4}x$ in the numerator and denominator to convert $\sin x$ into $\tan x$ by using the formula below:
Formula used:
$\dfrac{{\operatorname{Sin} x}}{{\operatorname{Cos} x}} = \tan x$
Now, we have to let t be equal to the $\tan x$ and then we have to differentiate this function with respect to t from both sides to get the $dx$ function into the $dt$ function.
Now, finally we have to integrate that $dt$ function with respect to t to obtain the desired answer.

Complete step by step answer:
First of all we have to let the given integration equal to I as mentioned below:
$I = \int {{{\operatorname{Sec} }^{4/3}}} x.\cos e{c^{8/3}}xdx$
Now, we convert $\sec x$ and $\cos ecx$ into the $\cos x$ and $\sin x$ respectively. With the help of the formula below:
Formula used:
$\sec x$= $\dfrac{1}{{\operatorname{Cos} x}}$and $\cos ecx$= $\dfrac{1}{{\sin x}}$
So, We get:
$ \Rightarrow I = \int {\dfrac{1}{{{{\cos }^{4/3}}x.{{\sin }^{8/3}}x}}} dx$
\[
   \Rightarrow I = \int {\dfrac{{{{\cos }^4}x}}{{{{\cos }^4}x({{\cos }^{4/3}}x.{{\sin }^{8/3}}x)}}} dx \\
   \Rightarrow I = \int {\dfrac{{{{\sec }^4}x}}{{\dfrac{{({{\cos }^{4/3}}x.{{\sin }^{8/3}}x)}}{{{{\cos }^4}x}}}}} dx \\
 \]Formula used:
$\dfrac{{\operatorname{Sin} x}}{{\operatorname{Cos} x}} = \tan x$
Now, we have simplify the denominator of the above expression and then convert $\dfrac{{\sin x}}{{\cos x}}$ into $\tan x$ with the help of the formula mentioned below:
Formula used:
$\dfrac{{\operatorname{Sin} x}}{{\operatorname{Cos} x}} = \tan x$
\[
   \Rightarrow I = \int {\dfrac{{{{\sec }^4}x}}{{\dfrac{{{{\sin }^{8/3}}x}}{{{{\cos }^{ - 4/3}}x{{\cos }^4}x}}}}} dx \\
   \Rightarrow I = \int {\dfrac{{{{\sec }^2}x.{{\sec }^2}x}}{{\dfrac{{{{\sin }^{8/3}}x}}{{{{\cos }^{8/3}}x}}}}} dx \\
   \Rightarrow I = \int {\dfrac{{{{\sec }^2}x.{{\sec }^2}x}}{{{{\tan }^{8/3}}x}}} dx \\
 \]
Now, we have to convert ${\sec ^2}x$ into ${\tan ^2}x$ with the help of the formula mentioned below.
Formula used:
\[{\sec ^2}x\]= $1 + {\tan ^2}x$
So, We get:
\[ \Rightarrow I = \int {\dfrac{{{{\sec }^2}x.(1 + {{\tan }^2}x)}}{{{{\tan }^{8/3}}x}}} dx\]………………………………………….(1)
Now, We have to let $\tan x = t$ and then differentiate this function with respect to t from both sides. We obtained this function as mentioned below:
$
   \Rightarrow \tan x = t \\
   \Rightarrow \dfrac{d}{{dt}}(\tan x) = \dfrac{d}{{dt}}(t) \\
   \Rightarrow {\sec ^2}x\dfrac{{dx}}{{dt}} = 1 \\
   \Rightarrow {\sec ^2}xdx = dt \\
 $
Now, Put these values in the above mentioned expression (1) in step 4.
\[
   \Rightarrow I = \int {\dfrac{{(1 + {\operatorname{t} ^2})}}{{{t^{8/3}}}}} dt \\
   \Rightarrow I = \int {(\dfrac{1}{{{t^{8/3}}}}} + \dfrac{{{t^2}}}{{{t^{8/3}}}})dt \\
   \Rightarrow I = \int {\dfrac{1}{{{t^{8/3}}}}} dt + \int {\dfrac{{{t^2}}}{{{t^{8/3}}}}dt} \\
   \Rightarrow I = \int {{t^{ - 8/3}}} dt + \int {{t^{ - 2/3}}dt} \\
    \\
    \\
 \]
Now, We have integrated the above expression with respect to t.
\[
   \Rightarrow I = \dfrac{{{t^{ - 8/3 + 1}}}}{{ - 8/3 + 1}} + \dfrac{{{t^{ - 2/3 + 1}}}}{{ - 2/3 + 1}} + C \\
   \Rightarrow I = \dfrac{{{t^{ - 5/3}}}}{{ - 5/3}} + \dfrac{{{t^{1/3}}}}{{1/3}} + C \\
   \Rightarrow I = - \dfrac{3}{5}{t^{ - 5/3}} + 3{t^{1/3}} + C....................(2) \\
    \\
 \]
Now, we have to put $t = \tan x$ in the expression (2)
\[ \Rightarrow I = - \dfrac{3}{5}{\tan ^{ - 5/3}}x + 3{\tan ^{1/3}}x + C\]

The integration of the function $\int {{{\operatorname{Sec} }^{4/3}}} x.\cos e{c^{8/3}}xdx$ is \[ - \dfrac{3}{5}{\tan ^{ - 5/3}}x + 3{\tan ^{1/3}}x + C\] .Therefore option (B) is correct.

Note: To obtain the solution of integration it is necessary that we have changed $\sec x$ and $\cos ecx$ into the $\cos x$ and $\sin x$ respectively with the help of the formulas which is mentioned in the solution hint.
To make the solution easy we have to let the terms of the given integration be some integer so that we can minimize the given integration but we have to determine the differentiation of the term we let.