Question

$\int {\sqrt {\dfrac{{\cos x - {{\cos }^3}x}}{{1 - {{\cos }^3}x}}} } dx \\ $
$
  A.\dfrac{{ - 2}}{3}{\sin ^{ - 1}}\left( {{{\cos }^{\dfrac{3}{2}}}x} \right) + c \\
  B.\dfrac{3}{2}{\sin ^{ - 1}}\left( {{{\cos }^{\dfrac{3}{2}}}x} \right) + c \\
  C.\dfrac{{ - 2}}{3}{\cos ^{ - 1}}\left( {{{\cos }^{\dfrac{3}{2}}}x} \right) + c \\
  D.\dfrac{3}{2}{\cos ^{ - 1}}\left( {{{\cos }^{\dfrac{3}{2}}}x} \right) + c \\
 $

Answer 1

Hint:
The given function is indefinite since there is no limit given. The indefinite integral of a function is a differentiable function F whose derivative is equal to the original function f. The first fundamental theorem of calculus allows definite integrals to be computed in terms of indefinite integrals.
In the given function, we will try to bring all the identity in the same form, and then we will use the trigonometric identity \[1 - {\cos ^2}\theta = {\sin ^2}\theta \] and we further reduce the function and then it is integrated.

Complete step by step solution:
Let us consider that $I = \int {\sqrt {\dfrac{{\cos x - {{\cos }^3}x}}{{1 - {{\cos }^3}x}}} } dx$
The above equation can also be written by taking $\cos x$ as common from both the terms in the numerator as
$
  I = \int {\sqrt {\dfrac{{\cos x - {{\cos }^3}x}}{{1 - {{\cos }^3}x}}} } dx \\
   = \int {\sqrt {\dfrac{{\cos x(1 - {{\cos }^2}x)}}{{1 - {{\cos }^3}x}}} } dx \\
   = \int {\sqrt {\dfrac{{{{\sin }^2}x\cos x}}{{1 - {{\cos }^3}x}}} } dx - - - - (i) \\
 $
Now, let us consider that $u = {\cos ^{\dfrac{3}{2}}}x - - - - (ii)$
Differentiate the equation (ii) with respect to x as:
$
 \dfrac{d}{{dx}}(u) = \dfrac{d}{{dx}}\left( {{{\cos }^{\dfrac{3}{2}}}x} \right) \\
 \dfrac{{du}}{{dx}} = \dfrac{3}{2}{\cos ^{\left( {\dfrac{3}{2} - 1} \right)}}\left( { - \sin x} \right) \\
 du = \dfrac{{ - 3\sin x\sqrt {\cos x} }}{2}dx \\
 dx = \dfrac{{ - 2}}{{3\sin x\sqrt {\cos x} }}du - - - - (iii) \\
 $
Substituting the value of ‘dx’ obtained in equation (iii) in the equation (i) we get,
$
 I = \int {\sqrt {\dfrac{{{{\sin }^2}x\cos x}}{{1 - {{\cos }^3}x}}} } dx \\
   = \int {\sqrt {\dfrac{{{{\sin }^2}x\cos x}}{{1 - {{\cos }^3}x}}} \times } \dfrac{{ - 2}}{{3\sin x\sqrt {\cos x} }}du \\
   = \dfrac{{ - 2}}{3}\int {\dfrac{{\sin x}}{{\sin x}}\sqrt {\dfrac{{\cos x}}{{\cos x\left( {1 - {{\cos }^3}x} \right)}}} } du - - - - (iv) \\
 $
Again, substitute the values of the equation (ii) in the equation (iv) we get
$
  I = \dfrac{{ - 2}}{3}\int {\dfrac{{\sin x}}{{\sin x}}\sqrt {\dfrac{{\cos x}}{{\cos x\left( {1 - {{\cos }^3}x} \right)}}} } du \\
   = \dfrac{{ - 2}}{3}\int {\sqrt {\dfrac{1}{{1 - {{\cos }^3}x}}} } \\
   = \dfrac{{ - 2}}{3}\int {\sqrt {\dfrac{1}{{1 - {u^2}}}} du} - - - - (v){\text{ }}\left[ {u = {{\cos }^{\dfrac{3}{2}}}x \Rightarrow {{\cos }^3}x = {u^2}} \right] \\
 $
Using the definite integral formula \[\int {\dfrac{1}{{\sqrt {1 - {a^2}} }} = {{\sin }^{ - 1}}a} \] in the equation (v) we get
$
  I = \dfrac{{ - 2}}{3}\int {\sqrt {\dfrac{1}{{1 - {u^2}}}} du} \\
   = \dfrac{{ - 2}}{3}\left( {{{\sin }^1}u} \right) + c - - - - (vi) \\
 $
Now, substitute the value of $ u $ from the equation (ii) we get
$
  I = \dfrac{{ - 2}}{3}\left( {{{\sin }^1}u} \right) + c \\
   = \dfrac{{ - 2}}{3}{\sin ^{ - 1}}\left( {{{\cos }^{\dfrac{3}{2}}}x} \right) + c \\
 $
Hence, $\int {\sqrt {\dfrac{{\cos x - {{\cos }^3}x}}{{1 - {{\cos }^3}x}}} } dx = \dfrac{{ - 2}}{3}{\sin ^{ - 1}}\left( {{{\cos }^{\dfrac{3}{2}}}x} \right) + c$

Option A is correct.

Note:
While substituting the real parameter of the question with the auxiliary parameter, one should be sure that it will not make the problem more complex. However, selecting an auxiliary parameter completely depends on the individual point of view.