Key Concepts and Formulas

• Trigonometric Identity: $\sin^2x + \cos^2x = 1$
• Trigonometric Ratios: $\tan x = \frac{\sin x}{\cos x}$, $\sec x = \frac{1}{\cos x}$
• Integration by Substitution: $\int f(g(x))g'(x) dx = \int f(u) du$, where $u = g(x)$
• Power Rule for Integration: $\int x^n dx = \frac{x^{n+1}}{n+1} + C$, where $n \neq -1$

Step-by-Step Solution

Step 1: Factor the denominator. We begin by factoring the denominator of the given integral: $I = \int {{{{{\sin }^2}x{{\cos }^2}x} \over {{{\left( {{{\sin }^5}x + {{\cos }^3}x{{\sin }^2}x + {{\sin }^3}x{{\cos }^2}x + {{\cos }^5}x} \right)}^2}}}} dx$ $= \int {{{{{\sin }^2}x{{\cos }^2}x} \over {{{\left( {{{\sin }^3}x\left( {{{\sin }^2}x + {{\cos }^2}x} \right) + {{\cos }^3}x\left( {{{\sin }^2}x + {{\cos }^2}x} \right)} \right)}^2}}}} dx$ We factor out $\sin^3 x$ and $\cos^3 x$ from the terms in the parenthesis. This allows us to use the identity $\sin^2 x + \cos^2 x = 1$.

Step 2: Simplify using the trigonometric identity. Using the identity $\sin^2x + \cos^2x = 1$, we simplify the expression: $= \int {{{{{\sin }^2}x{{\cos }^2}x} \over {{{\left( {{{\sin }^3}x + {{\cos }^3}x} \right)}^2}}}} dx$

Step 3: Factor out $\cos^3 x$ from the denominator. Factor out $\cos^3 x$ from the term inside the parenthesis in the denominator to get $\tan x$ terms: $= \int {{{{{\sin }^2}x{{\cos }^2}x} \over {{{\left( {{{\cos }^3}x\left( {{{\tan }^3}x + 1} \right)} \right)}^2}}}} dx$ $= \int {{{{{\sin }^2}x{{\cos }^2}x} \over {{{\cos }^6}x{{\left( {{{\tan }^3}x + 1} \right)}^2}}}} dx$

Step 4: Simplify the expression. $= \int {{{{{\sin }^2}x} \over {{{\cos }^4}x{{\left( {{{\tan }^3}x + 1} \right)}^2}}}} dx$ $= \int {{{{{\sin }^2}x} \over {{{\cos }^2}x}} \cdot {1 \over {{{\cos }^2}x}} \cdot {1 \over {{{\left( {{{\tan }^3}x + 1} \right)}^2}}}} dx$ $= \int {{{\tan }^2}x{{\sec }^2}x{{\left( {1 + {{\tan }^3}x} \right)}^{ - 2}}} dx$

Step 5: Apply u-substitution. Let $u = 1 + \tan^3 x$. Then, $\frac{du}{dx} = 3\tan^2 x \sec^2 x$, which implies $du = 3\tan^2 x \sec^2 x dx$. $I = \int {{{\tan }^2}x{{\sec }^2}x{{\left( {1 + {{\tan }^3}x} \right)}^{ - 2}}} dx = \frac{1}{3} \int {3{{\tan }^2}x{{\sec }^2}x{{\left( {1 + {{\tan }^3}x} \right)}^{ - 2}}} dx$ $= \frac{1}{3}\int {{u^{ - 2}}} du$

Step 6: Integrate with respect to u. Using the power rule for integration: $= \frac{1}{3} \cdot \frac{{{u^{ - 1}}}}{{ - 1}} + C$ $= - \frac{1}{{3u}} + C$

Step 7: Substitute back for x. Replace $u$ with $1 + \tan^3 x$: $= - \frac{1}{{3\left( {1 + {{\tan }^3}x} \right)}} + C$

Step 8: Rewrite in terms of cotangent. Since the correct answer is in terms of $\cot x$, we can rewrite our answer. We know that $\tan x = \frac{1}{\cot x}$, so $\tan^3 x = \frac{1}{\cot^3 x}$. Substituting this: $- \frac{1}{{3\left( {1 + \frac{1}{{{{\cot }^3}x}}} \right)}} + C = - \frac{1}{{3\left( {\frac{{{{\cot }^3}x + 1}}{{{{\cot }^3}x}}} \right)}} + C = - \frac{{{{\cot }^3}x}}{{3\left( {1 + {{\cot }^3}x} \right)}} + C$

However, the provided answer is $\frac{-1}{1 + \cot^3 x} + C$. Let's manipulate this to see if they are equivalent: $\frac{-1}{1 + \cot^3 x} + C = \frac{-1}{1 + \frac{1}{\tan^3 x}} + C = \frac{-\tan^3 x}{\tan^3 x + 1} + C = \frac{-((\tan^3 x + 1) - 1)}{\tan^3 x + 1} + C = \frac{-(\tan^3 x + 1)}{\tan^3 x + 1} + \frac{1}{\tan^3 x + 1} + C = -1 + \frac{1}{\tan^3 x + 1} + C = \frac{1}{1 + \tan^3 x} + C'$ where $C' = C - 1$.

Going back to our solution, we made an error in the problem statement. We should have arrived at $= {{ - 1} \over {3\left( {1 + {{\tan }^3}x} \right)}} + c$

But the correct answer is ${{ - 1} \over {1 + {{\cot }^3}x}} + C$. So our step 8 is wrong. Let us try to work backwards from the answer. $= {{ - 1} \over {1 + {{\cot }^3}x}} + C = {{- 1} \over {1 + \frac{1}{\tan^3 x}}} + C = {{- 1} \over {\frac{\tan^3 x + 1}{\tan^3 x}}} + C = \frac{-\tan^3 x}{1 + \tan^3 x} + C = \frac{-(1 + \tan^3 x) + 1}{1 + \tan^3 x} + C = -1 + \frac{1}{1 + \tan^3 x} + C = \frac{1}{1 + \tan^3 x} + C'$ where $C' = C - 1$.

Let's reconsider our u-substitution. $u = \tan x$, then $du = \sec^2 x dx$. $I = \int \frac{\tan^2 x \sec^2 x}{(1 + \tan^3 x)^2} dx = \int \frac{u^2}{(1 + u^3)^2} du$ Let $v = 1 + u^3$, then $dv = 3u^2 du$. $I = \frac{1}{3} \int \frac{3u^2}{(1 + u^3)^2} du = \frac{1}{3} \int \frac{dv}{v^2} = \frac{1}{3} \int v^{-2} dv = \frac{1}{3} \frac{v^{-1}}{-1} + C = \frac{-1}{3v} + C = \frac{-1}{3(1 + \tan^3 x)} + C$ We still arrive at the same answer.

$\frac{-1}{3(1 + \tan^3 x)} = \frac{-1}{3(1 + \frac{1}{\cot^3 x})} = \frac{-1}{3(\frac{\cot^3 x + 1}{\cot^3 x})} = \frac{-\cot^3 x}{3(1 + \cot^3 x)}$ There seems to be an error in the given correct answer.

Step 9: Correct the error, we should arrive at option (C) The correct answer should be $= {{ - 1} \over {3\left( {1 + {{\tan }^3}x} \right)}} + C$

Common Mistakes & Tips

• Remember to include the constant of integration, $C$, after performing indefinite integration.
• Be careful when substituting back after u-substitution. Ensure you're substituting correctly.
• When simplifying trigonometric expressions, keep the target expression in mind to guide your steps.

Summary

We factored the denominator, used trigonometric identities to simplify the integral, and then applied u-substitution to solve the integral. We then arrived at the answer in terms of $\tan x$.

Final Answer

The final answer is \boxed{{{- 1} \over {3\left( {1 + {{\tan }^3}x} \right)}} + C}, which corresponds to option (C).