Question

he integral\[\int {{{\sec }^{\dfrac{2}{3}}}} x\cos e{c^{\dfrac{4}{3}}}xdx\] is equal to (Here C is constant of integration).
$
  {\text{A}}{\text{. }}3{\tan ^{ - 1/3}}x + C \\
  {\text{B}}{\text{. }} - \dfrac{3}{4}{\tan ^{ - 4/3}}x + C \\
  {\text{C}}{\text{. }} - 3{\cot ^{ - 1/3}}x + C \\
  {\text{D}}.{\text{ }} - 3{\tan ^{ - 1/3}}x + C \\
$

Answer 1

Hint- In mathematics, an integral assigns numbers of functions in a way that can describe displacement area, volume, and other concepts that arise by combining infinitesimal sections. Its inverse operation is differentiation.
\[\int_a^b {f(x)dx} = [F(x)]_a^b = F(b) - F(a)\]
Integration is a way of adding slices to find the whole. Integration can be used to find areas, volumes, central points, and many useful things. Some of the useful trigonometric identities which are going to be used here are:
\[
  \sec x = \dfrac{1}{{\cos x}} \\
  \cos ecx = \dfrac{1}{{\sin x}} \\
 \]


Complete step by step solution:
Converting \[\sec x\] and \[{\text{cosec }}x\] according to the above we get:
\[
  I = \int {{{\sec }^{\dfrac{2}{3}}}} x\cos e{c^{\dfrac{4}{3}}}xdx \\
   = \int {\dfrac{{dx}}{{{{(\cos x)}^{\dfrac{2}{3}}}{{(\sin x)}^{\dfrac{4}{3}}}}}} \\
 \]
Now simplifying again,
\[
  I = \int {\dfrac{{\left( {\dfrac{1}{{{{\cos }^2}x}}} \right)}}{{^{\left( {{{\left( {\dfrac{{\sin x}}{{\cos x}}} \right)}^{\dfrac{4}{3}}}} \right)}}}} dx \\
   = \int {\dfrac{{{{\sec }^2}x}}{{{{(\tan )}^{\dfrac{4}{3}}}}}} dx - - - - (i) \\
 \]
Let, \[\tan x = t \Rightarrow {\sec ^2}xdx = dt\]
Substitute \[\tan x = t \Rightarrow {\sec ^2}xdx = dt\] in equation (i) we get,
\[
  I = \int {\dfrac{{dt}}{{{t^{\dfrac{4}{3}}}}}} \\
   = \int {{t^{\dfrac{{ - 4}}{3}}}dt} \\
 \]
Applying the integral property, \[\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} \] we get:
\[
  I = \dfrac{{{t^{\dfrac{{ - 4}}{3} + 1}}}}{{\left( {^{\dfrac{{ - 4}}{3} + 1}} \right)}} \\
  I = \dfrac{{{t^{\dfrac{{ - 1}}{3}}}}}{{\left( {\dfrac{{ - 1}}{3}} \right)}} \\
  I = \dfrac{{ - 3}}{{\left( {{t^{\dfrac{1}{3}}}} \right)}} + C \\
  I = \dfrac{{ - 3}}{{{{(\tan )}^{\dfrac{1}{3}}}}} + C \\
  I = - 3{(\tan )^{\dfrac{{ - 1}}{3}}} + C \\
 \]
Hence, \[\int {{{\sec }^{\dfrac{2}{3}}}} x\cos e{c^{\dfrac{4}{3}}}xdx = - 3{(\tan )^{\dfrac{{ - 1}}{3}}} + C\]
Option D is correct.


Note: The product of integer power of trigonometric functions is the product of integer power of cosine and sine functions. Thus, one can evaluate the integration of the product of integer power of trigonometric functions by applying the reduction formula of integration of the product of integer power of cosine and sine functions. In general, we can evaluate the integration of the product of the integer power of trigonometric functions.
In calculation, integration substitution, also known as u-substitution or change of variables, is a method of evaluating integral. Direct application of the fundamental theorem of calculus to find an antiderivative can be quite difficult, and integration by simplifying that task.
This question also is done by using the formulae of trigonometry.