Question

Find the value of integral:
\[\int{\dfrac{{{\sin }^{2}}x.{{\cos }^{2}}x}{{{\left( {{\sin }^{3}}x+{{\cos }^{3}}x \right)}^{2}}}dx}\]
(a) \[\dfrac{1}{\left( 1+{{\cot }^{2}}x \right)}+c\]
(b) \[-\dfrac{1}{3\left( 1+{{\tan }^{3}}x \right)}+c\]
(c) \[\dfrac{{{\sin }^{3}}x}{1+{{\cos }^{3}}x}+c\]
(d) \[\dfrac{{{\cos }^{3}}x}{3\left( 1+{{\sin }^{3}}x \right)}+c\]

Answer 1

Hint: Convert the sin(x) and cos(x) into in to tan(x) and sec(x) and apply power rule:
\[\int{{{v}^{n}}dv}=\dfrac{{{v}^{n+1}}}{n+1}\]
We need to use the method of integration by substitution twice in order to get the value of the given integral. After solving the integral remember to put the original values back into the final equation.

Complete step-by-step answer:
Given expression in the question, for which integral is asked:
\[\int{\dfrac{{{\sin }^{2}}x.{{\cos }^{2}}x}{{{\left( {{\sin }^{3}}x+{{\cos }^{3}}x \right)}^{2}}}dx}\]
We know:
sin (x) = tan (x) . cos (x)
We also know that:
\[\cos x=\dfrac{1}{\sec x}\]
If we substitute cos (x) into first equation we get:
sin (x) = \[\dfrac{\tan x}{\sec x}\]
By substituting the values of sin (x) and cos (x) into the given expression it converts into:
\[\int{\dfrac{\left( \dfrac{{{\tan }^{2}}x}{{{\sec }^{2}}x} \right)\left( \dfrac{1}{{{\sec }^{2}}x} \right)}{{{\left( \left( \dfrac{{{\tan }^{3}}x}{{{\sec }^{3}}x} \right)+\left( \dfrac{1}{{{\sec }^{3}}x} \right) \right)}^{2}}}}dx\]
By taking least common multiple (L.C.M.) in denominator, we get:
\[\int{\dfrac{\left( \dfrac{{{\tan }^{2}}x}{{{\sec }^{2}}x} \right)\left( \dfrac{1}{{{\sec }^{2}}x} \right)}{{{\left( \dfrac{{{\tan }^{3}}x+1}{{{\sec }^{3}}x} \right)}^{2}}}}dx\]
By bringing the sec (x) term into numerator, we get:
\[\int{\left( \dfrac{{{\tan }^{2}}x}{{{\sec }^{2}}x} \right)\left( \dfrac{1}{{{\sec }^{2}}x} \right)\left( \dfrac{{{\sec }^{6}}x}{{{\left( {{\tan }^{3}}x+1 \right)}^{2}}} \right)dx}\]
By cancelling the common terms, we get:
\[\int{\dfrac{{{\tan }^{2}}x.{{\sec }^{2}}x}{{{\left( {{\tan }^{3}}x+1 \right)}^{2}}}dx}\]
Now by substitution method convert the terms in to terms of u:
Substitute u = tan (x)
We need to find du:
du = d(tan (x))
\[du={{\sec }^{2}}x.dx\]
By substituting above values, we get:
\[\int{\dfrac{{{u}^{2}}}{\left( {{u}^{3}}+1 \right)}du}\]
Now use the general algebraic identity:
\[{{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)\]
Here in our case:
a = u,
b = 1,
Now by applying above formula, we get:
\[{{u}^{3}}+1=\left( u+1 \right)\left( {{u}^{2}}-u+1 \right)\]
By substituting this in to the integration, we get:
\[\int{\dfrac{{{u}^{2}}}{{{\left( u+1 \right)}^{2}}{{\left( {{u}^{2}}-u+1 \right)}^{2}}}du}\]
Now again use the substitution method.
Now we will substitute:
v = \[\left( u+1 \right)\left( {{u}^{2}}-u+1 \right)\]
Now we need to find the value of dv for the substitution.
\[dv=d\left( \left( u+1 \right)\left( {{u}^{2}}-u+1 \right) \right)\]
By using formula:
\[d\left( a.b \right)=a.db+b.da\]
By simplification, we get:
\[dv=\left( {{u}^{2}}+\left( u+1 \right)\left( 2u-1 \right)-u+1 \right)du\]
By cancelling the common terms, we get:
\[\begin{align}
  & dv=3.{{u}^{2}}du \\
 & {{u}^{2}}du=\dfrac{dv}{3} \\
\end{align}\]
By substituting the values of v, dv into integration, we get:
\[\int{\dfrac{1}{{{v}^{2}}}\dfrac{dv}{3}}=\dfrac{1}{3}\int{\dfrac{1}{{{v}^{2}}}}\]
Now by using the power formula:
\[\int{{{v}^{n}}dv}=\dfrac{{{v}^{n+1}}}{n+1}\]
Integration can be written as:
\[\dfrac{1}{3}\int{{{v}^{-2}}dv}\]
Now by applying power formula, with n = -2, we get:
\[\dfrac{1}{3}\dfrac{{{v}^{-1}}}{-1}=-\dfrac{1}{3v}\]
Now substitute the v value back into the expression:
v = \[\left( u+1 \right)\left( {{u}^{2}}-u+1 \right)\] By doing this, we get:
\[-\dfrac{1}{3\left( u+1 \right)\left( {{u}^{2}}-u+1 \right)}\]
Now by substituting the u value back into expression:
u = tan (x)
By doing this, we get:
\[-\dfrac{1}{3\left( \tan x+1 \right)\left( {{\tan }^{2}}x-\tan x+1 \right)}\]
Now use the algebraic identity: \[{{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)\]
a = tan (x), b = 1,
By using this, we get:
\[-\dfrac{1}{3\left( 1+{{\tan }^{3}}x \right)}\]
Therefore by above solving we can say:
\[\]
So option (b) is correct.

Note: Alternate method: Directly take \[{{\tan }^{3}}x+1=v\] as substitution before taking u.
This is a tough idea to get. But using this, the question will be solved easily.