Question

Solve the following:
\[\int {\dfrac{{{{\sin }^3}xdx}}{{\left( {1 + {{\cos }^2}x} \right)\sqrt {1 + {{\cos }^2}x + {{\cos }^4}x} }}} \]
A.\[{\sec ^{ - 1}}\left( {\sec x + \cos x} \right) + C\]
B.\[{\sec ^{ - 1}}\left( {\sec x - \cos x} \right) + C\]
C.\[{\sec ^{ - 1}}\left( {\cos x - \tan x} \right) + C\]
D.\[{\sec ^{ - 1}}\left( {\cos x + \tan x} \right) + C\]

Answer 1

Hint: In the given question , we have options given in \[{\sec ^{ - 1}}\] , so we have to use the formula of the derivative of the \[{\sec ^{ - 1}}\] which is \[\dfrac{{d\left( {{{\sec }^{ - 1}}x} \right)}}{{dx}} = \dfrac{1}{{x\sqrt {{x^2} - 1} }}\] . First we simplify the given expression in terms \[\sec x\] , then integrate the expression accordingly .

Complete step-by-step answer:
Given : \[\int {\dfrac{{{{\sin }^3}xdx}}{{\left( {1 + {{\cos }^2}x} \right)\sqrt {1 + {{\cos }^2}x + {{\cos }^4}x} }}} \]
Now , in the denominator we will take \[\cos x\] and \[{\cos ^2}x\] from the under root term , we get
\[\int {\dfrac{{{{\sin }^3}xdx}}{{\cos x\left( {\sec x + \cos x} \right)\cos x\sqrt {{{\sec }^2}x + 1 + {{\cos }^2}x} }}} \]
Now adding and subtracting \[1\] in the under root term of denominator ,
\[\int {\dfrac{{{{\sin }^3}xdx}}{{\cos x\left( {\sec x + \cos x} \right)\cos x\sqrt {{{\sec }^2}x + 1 + 1 - 1+{{\cos }^2}x} }}} \]
On simplifying we get ,
\[\int {\dfrac{{{{\sin }^3}xdx}}{{\cos x\left( {\sec x + \cos x} \right)\cos x\sqrt {{{\sec }^2}x + 2 + {{\cos }^2}x - 1} }}} \]
Now using the identity of \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\] in denominator we get ,
\[\int {\dfrac{{{{\sin }^3}xdx}}{{{{\cos }^2}x\left( {\sec x + \cos x} \right)\sqrt {{{\left( {\sec x + \cos x} \right)}^2} - 1} }}} \]
Now let \[\left( {\sec x + \cos x} \right) = t\]
On simplification we get ,
\[\dfrac{1}{{\cos x}} + \cos = t\]
On simplifying we get ,
\[\dfrac{{1 + {{\cos }^2}x}}{{\cos x}} = t\]
Now differentiating w.r.t \[x\] using quotient rule , we get
\[dt = \dfrac{{\left( { - 2\cos x\sin x} \right)\cos x - \left\{ { - \sin x\left( {1 + {{\cos }^2}x} \right)} \right\}}}{{{{\cos }^2}x}}dx\]
On simplifying we get ,
\[dt = \dfrac{{\left( { - 2\cos x\sin x} \right)\cos x + \sin x\left( {1 + {{\cos }^2}x} \right)}}{{{{\cos }^2}x}}dx\]
On solving we get ,
\[dt = \dfrac{{ - 2{{\cos }^2}x\sin x + \sin x{{\cos }^2}x + \sin x}}{{{{\cos }^2}x}}dx\]
On simplifying we get ,
\[dt = \dfrac{{ - {{\cos }^2}x\sin x + \sin x}}{{{{\cos }^2}x}}dx\]
Taking \[\sin x\] common we get ,
\[dt = \dfrac{{\sin x\left( {1 - {{\cos }^2}x} \right)}}{{{{\cos }^2}x}}dx\]
On using the trigonometric identity \[{\sin ^2}x + {\cos ^2}x = 1\] we get ,
\[dt = \dfrac{{\sin x\left( {{{\sin }^2}x} \right)}}{{{{\cos }^2}x}}dx\]
On simplifying we get ,
\[dt = \dfrac{{{{\sin }^3}x}}{{{{\cos }^2}x}}dx\]
Now putting the value of \[dt\] and \[\left( {\sec x + \cos x} \right) = t\] we get ,
\[\int {\dfrac{{dt}}{{t\sqrt {{t^2} - 1} }}} \]
Now we know that derivative of \[{\sec ^{ - 1}}x\] is \[\dfrac{1}{{x\sqrt {{x^2} - 1} }}\], so on integrating we will get \[{\sec ^{ - 1}}x\] .
On integrating we get ,
\[ = {\sec ^{ - 1}}t + C\]
Now we will put the value of \[t\] we get ,
\[ = {\sec ^{ - 1}}\left( {\sec x + \cos x} \right) + C\]
So, the correct answer is “Option A”.

Note: When you make substitution always do it in such a way that the derivative of that will get adjusted in the given expression and then integrate accordingly . Also , write the value which you have substituted or let . In the final answer write \[C\] ( constant ) , as it makes the answer complete .