Key Concepts and Formulas

• Inverse Trigonometric Identities: $\cot^{-1}(y) = \tan^{-1}(1/y)$ for $y > 0$.
• Tangent Addition Formula: $\tan^{-1}A + \tan^{-1}B = \tan^{-1}\left(\frac{A+B}{1-AB}\right)$, provided $AB < 1$.
• Definite Integral Property (King's Rule): $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$.
• Integration by Parts: $\int u \, dv = uv - \int v \, du$.
• Standard Integral: $\int \tan^{-1}(x) dx = x\tan^{-1}(x) - \frac{1}{2}\log_e(1+x^2) + C$.

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Step-by-Step Solution:

Let the given integral be $I$. $I = \int\limits_0^1 {x{{\cot }^{ - 1}}(1 - {x^2} + {x^4})dx}$

Step 1: Substitution to simplify the integrand

The argument of the $\cot^{-1}$ function involves $x^2$ and $x^4$, and there is an $x$ multiplying the inverse trigonometric term. This suggests a substitution $t = x^2$. Differentiating with respect to $x$, we get $dt = 2x \, dx$, which implies $x \, dx = \frac{1}{2} dt$. We also need to change the limits of integration: When $x=0$, $t = 0^2 = 0$. When $x=1$, $t = 1^2 = 1$. Substituting these into the integral: $I = \int\limits_0^1 {{\cot }^{ - 1}}(1 - t + t^2) \left(\frac{1}{2} dt\right)$ $I = \frac{1}{2}\int\limits_0^1 {{{\cot }^{ - 1}}\left( {1 - t + {t^2}} \right)dt}$

• Why this step? The substitution $t=x^2$ simplifies the argument of the inverse trigonometric function and converts the $x \, dx$ term into a more manageable $\frac{1}{2} dt$ term, making the integration process easier.

Step 2: Convert $\cot^{-1}$ to $\tan^{-1}$

We use the identity $\cot^{-1}(y) = \tan^{-1}(1/y)$ for $y > 0$. The argument is $y = 1 - t + t^2$. We can rewrite this as $y = (t - 1/2)^2 + 3/4$. For $t \in [0,1]$, $(t - 1/2)^2 \ge 0$, so $y \ge 3/4$. Thus, $y > 0$, and the identity can be applied. ${\cot }^{ - 1}\left( {1 - t + {t^2}} \right) = {\tan }^{ - 1}\left( \frac{1}{1 - t + {t^2}} \right)$ The integral becomes: $I = \frac{1}{2}\int\limits_0^1 {{{\tan }^{ - 1}}\left( {\frac{1}{1 - t + {t^2}}} \right)dt}$

• Why this step? The $\tan^{-1}$ function is often easier to work with, especially in conjunction with the tangent addition formula, which is a common technique for simplifying such integrals.

Step 3: Apply the tangent addition formula

We aim to express $\frac{1}{1 - t + t^2}$ in the form $\frac{A+B}{1-AB}$ to use the identity $\tan^{-1}A + \tan^{-1}B = \tan^{-1}\left(\frac{A+B}{1-AB}\right)$. Let $A=t$ and $B=1-t$. Then, $A+B = t + (1-t) = 1$. And $AB = t(1-t) = t - t^2$. So, the denominator $1 - t + t^2$ can be written as $1 - (t - t^2) = 1 - t(1-t)$. Thus, $\frac{1}{1 - t + t^2} = \frac{t + (1-t)}{1 - t(1-t)}$. For $t \in [0,1]$, $t(1-t) \le 1/4 < 1$, so the condition $AB < 1$ for the tangent addition formula is satisfied. Therefore, ${\tan }^{ - 1}\left( \frac{1}{1 - t + {t^2}} \right) = {\tan }^{ - 1}(t) + {\tan }^{ - 1}(1-t)$ Substituting this back into the integral: $I = \frac{1}{2}\int\limits_0^1 {\left[ {{{\tan }^{ - 1}}(t) + {{\tan }^{ - 1}}(1 - t)} \right]dt}$

• Why this step? This is a crucial algebraic manipulation. By decomposing the $\tan^{-1}$ term into a sum of two simpler $\tan^{-1}$ terms, we make the integral amenable to further simplification using integral properties.

Step 4: Utilize the King's Rule for Definite Integrals

We can split the integral into two parts: $I = \frac{1}{2}\int\limits_0^1 {{{\tan }^{ - 1}}(t)dt} + \frac{1}{2}\int\limits_0^1 {{{\tan }^{ - 1}}(1 - t)dt}$ Now, let's focus on the second integral, $\int\limits_0^1 {{{\tan }^{ - 1}}(1 - t)dt}$. We apply the King's Rule: $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$. Here, $a=0$, $b=1$, and the variable is $t$. So, $a+b-t = 0+1-t = 1-t$. Applying this rule to the second integral: $\int\limits_0^1 {{{\tan }^{ - 1}}(1 - t)dt} = \int\limits_0^1 {{{\tan }^{ - 1}}(1 - (1 - t))dt} = \int\limits_0^1 {{{\tan }^{ - 1}}(t)dt}$ Substituting this result back into the expression for $I$: $I = \frac{1}{2}\int\limits_0^1 {{{\tan }^{ - 1}}(t)dt} + \frac{1}{2}\int\limits_0^1 {{{\tan }^{ - 1}}(t)dt}$ $I = 2 \times \frac{1}{2}\int\limits_0^1 {{{\tan }^{ - 1}}(t)dt}$ $I = \int\limits_0^1 {{{\tan }^{ - 1}}(t)dt}$

• Why this step? The King's Rule is exceptionally useful for integrals with symmetric limits like $[0,1]$. It allows us to transform one part of the integral into the other, effectively doubling the common integral and simplifying the problem significantly.

Step 5: Evaluate the integral using Integration by Parts

We need to evaluate $\int\limits_0^1 {{{\tan }^{ - 1}}(t)dt}$. We use integration by parts with $u = \tan^{-1}(t)$ and $dv = dt$. Then, $du = \frac{1}{1+t^2} dt$ and $v = t$. Applying the integration by parts formula: $\int\limits_0^1 {{{\tan }^{ - 1}}(t)dt} = \left[ t{\tan }^{ - 1}(t) \right]_0^1 - \int\limits_0^1 {t \cdot \frac{1}{1+t^2} dt}$ Evaluate the first term: $\left[ t{\tan }^{ - 1}(t) \right]_0^1 = (1 \cdot \tan^{-1}(1)) - (0 \cdot \tan^{-1}(0)) = 1 \cdot \frac{\pi}{4} - 0 = \frac{\pi}{4}$ Now, evaluate the integral term: $\int\limits_0^1 {\frac{t}{1+t^2} dt}$. Let $u = 1+t^2$. Then $du = 2t \, dt$, so $t \, dt = \frac{1}{2} du$. The limits of integration also change: When $t=0$, $u = 1+0^2 = 1$. When $t=1$, $u = 1+1^2 = 2$. So, the integral becomes: $\int\limits_1^2 {\frac{1}{u} \left(\frac{1}{2} du\right)} = \frac{1}{2}\int\limits_1^2 {\frac{1}{u} du} = \frac{1}{2} \left[ \log_e|u| \right]_1^2$ $= \frac{1}{2} (\log_e 2 - \log_e 1) = \frac{1}{2} (\log_e 2 - 0) = \frac{1}{2} \log_e 2$ Combining the results from integration by parts: $I = \frac{\pi}{4} - \frac{1}{2} \log_e 2$

• Why this step? Integration by parts is a standard technique for integrating functions like $\tan^{-1}(x)$. The resulting integral $\int \frac{t}{1+t^2} dt$ is a simple logarithmic integral, which can be solved using a basic substitution.

Common Mistakes & Tips

• Algebraic Errors: Carefully check the algebraic manipulations, especially when applying the tangent addition formula. A small mistake here can lead to a completely wrong answer.
• Substitution Limits: Always remember to change the limits of integration when performing a substitution in a definite integral.
• Inverse Trig Identities: Ensure the conditions for applying inverse trigonometric identities (like $y>0$ for $\cot^{-1}(y) = \tan^{-1}(1/y)$) are met.

Summary

The problem was solved by first simplifying the integral using a substitution $t=x^2$. Then, the $\cot^{-1}$ function was converted to $\tan^{-1}$ to enable the application of the tangent addition formula, which decomposed the integrand into a sum of two $\tan^{-1}$ terms. The King's Rule was then used to simplify the integral to a single $\tan^{-1}(t)$ integral. Finally, integration by parts was employed to evaluate this integral, yielding the result.

The final answer is $\boxed{{\pi \over 4} - {1 \over 2}{\log _e}2}$.