Question

The integral $\int {\dfrac{{{{\sin }^2}x{{\cos }^2}x}}{{{{\left( {{{\sin }^5}x + {{\cos }^3}x{{\sin }^2}x + {{\sin }^3}x{{\cos }^2}x + {{\cos }^5}x} \right)}^2}}}dx}$ is equal to
A) $\dfrac{1}{{1 + {{\cot }^3}x}} + C$
B) $\dfrac{{ - 1}}{{1 + {{\cot }^3}x}} + C$
C) $\dfrac{1}{{3\left( {1 + {{\tan }^3}x} \right)}} + C$
D) $\dfrac{{ - 1}}{{3\left( {1 + {{\tan }^3}x} \right)}} + C$

Answer 1

At first, we need to simplify the integral. In order to do that, take ${\sin ^2}x$ and ${\cos ^2}x$ in such a way that we get ${\sin ^3}x + {\cos ^3}x$ in the denominator. Then, divide the numerator and denominator by ${\cos ^6}x$ and find the integral.

Complete step-by-step solution:
Let us assume that
$I = \int {\dfrac{{{{\sin }^2}x{{\cos }^2}x}}{{{{\left( {{{\sin }^5}x + {{\cos }^3}x{{\sin }^2}x + {{\sin }^3}x{{\cos }^2}x + {{\cos }^5}x} \right)}^2}}}dx}$
Now, we have to simplify the denominator. But how will we do that? We will take ${\sin ^2}x$ common from the first two terms and ${\cos ^2}x$ common from the last two terms.
$I = \int {\dfrac{{{{\sin }^2}x{{\cos }^2}x}}{{{{\left( {{{\sin }^5}x + {{\cos }^3}x{{\sin }^2}x + {{\sin }^3}x{{\cos }^2}x + {{\cos }^5}x} \right)}^2}}}dx} \\\ = \int {\dfrac{{{{\sin }^2}x{{\cos }^2}x}}{{{{\left( {{{\sin }^2}x\left( {{{\sin }^3}x + {{\cos }^3}x} \right) + {{\cos }^2}x\left( {{{\sin }^3}x + {{\cos }^3}x} \right)} \right)}^2}}}dx} \\\ = \int {\dfrac{{{{\sin }^2}x{{\cos }^2}x}}{{{{\left( {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)\left( {{{\sin }^3}x + {{\cos }^3}x} \right)} \right)}^2}}}dx}$
Using the property, ${\sin ^2}x + {\cos ^2}x = 1$
$I = \int {\dfrac{{{{\sin }^2}x{{\cos }^2}x}}{{{{\left( {{{\sin }^3}x + {{\cos }^3}x} \right)}^2}}}dx}$..................(1)
Now we will divide both numerator and denominator by ${\cos ^6}x$, i.e., power of $\cos x$ in denominator. But do we need to do this?
This is because solving such an integral is quite difficult and dividing it by ${\cos ^6}x$ will give us terms in $\tan x$ and $\sec x$.
Hence, dividing both numerator and denominator by ${\cos ^6}x$ in equation (1)
$I = \int {\dfrac{{\dfrac{{{{\sin }^2}x{{\cos }^2}x}}{{{{\cos }^6}x}}}}{{\dfrac{{{{\left( {{{\sin }^3}x + {{\cos }^3}x} \right)}^2}}}{{{{\cos }^6}x}}}}dx} \\\ = \int {\dfrac{{\dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}} \times \dfrac{{{{\cos }^2}x}}{{{{\cos }^4}x}}}}{{{{\left( {\dfrac{{{{\sin }^3}x + {{\cos }^3}x}}{{{{\cos }^3}x}}} \right)}^2}}}dx} \\\ = \int {\dfrac{{{{\tan }^2}x{{\sec }^2}x}}{{{{\left( {{{\tan }^3}x + 1} \right)}^2}}}dx}$
Now, we can solve this by substitution.
Let $t = {\tan ^3}x + 1$.......................(2)
Then, $dt = 3{\tan ^2}x{\sec ^2}xdx$
$\dfrac{{dt}}{3} = {\tan ^2}x{\sec ^2}xdx$.......................(3)
Using (2) and (3) in $I$,
$I = \int {\dfrac{1}{{3{t^2}}}dt} \\\ {\text{Now, }}\int {\dfrac{1}{{{t^2}}}dt} = - \dfrac{1}{t} + C \\\ \Rightarrow I = - \dfrac{1}{{3t}} + C$
Replacing $t$ by ${\tan ^3}x + 1$
$I = - \dfrac{1}{{3\left( {{{\tan }^3}x + 1} \right)}} + C$

Hence, option D is correct.

Note: In the first step, we can also take ${\sin ^3}x$ common from first and third term and ${\cos ^3}x$ common from second and fourth term. Dividing by ${\cos ^6}x$, that is, the power of the $\cos x$ in the denominator is the key to solve this question. Choosing the correct substitution is also very important.