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 .