Question

If \(\int\left[\left\\{x^{2012} \tan ^{-1}\left(x^{2012}\right)\right\\} /\left(1+x^{4024}\right)\right] d x=k \tan ^{-1}\left(x^{2012}\right)+c\) (a) \([1 /(2012)]\) (a) \(-[1 /(2012)]\) (c) \([1 /(4024)]\) (d) \(-[1 /(4024)]\)

Short answer

(a) \([1/(2012)]\)

Step-by-step solution

Recognize the functions for integration by parts

Integration by parts formula is: \[\int u \, dv = uv - \int v \, du\] Here, we can choose the two functions to apply the integration by parts formula: Let u = \(\tan^{-1}(x^{2012})\) Let dv = \(\frac{x^{2012}}{1+x^{4024}}dx\)

Calculate du and v

We need to find du and v for using integration by parts formula. Derivative of u = du: \[du = \frac{d}{dx}(\tan^{-1}(x^{2012}))\cdot dx = \frac{2012x^{2011}}{1+x^{4024}}dx\] Now for v, we have: \(v = \int \frac{x^{2012}}{1+x^{4024}}dx\) Notice that the denominator is a perfect square: \((1+x^{2012})^2 = 1+x^{4024}\). Let's substitute: \(x^{2012} = y\), then \(2012x^{2011}dx = dy\). We can see that our integral now becomes: \[v=\int\frac{1}{y^2+1}dy\] This integral is the derivative of the inverse tangent function, so we get: \[v = \tan^{-1}(y) = \tan^{-1}(x^{2012})\]

Apply the integration by parts formula

Now we can apply the integration by parts formula: \[\int \frac{x^{2012} \tan^{-1}(x^{2012})}{1+x^{4024}} dx = uv - \int v du\] \[\int \frac{x^{2012} \tan^{-1}(x^{2012})}{1+x^{4024}} dx = (\tan^{-1}(x^{2012}))(\tan^{-1}(x^{2012})) - \int (\tan^{-1}(x^{2012}))(\frac{2012x^{2011}}{1+x^{4024}}dx)\] Notice that we get back the same term we started with on the right side: \[\int \frac{x^{2012} \tan^{-1}(x^{2012})}{1+x^{4024}} dx = (\tan^{-1}(x^{2012}))(\tan^{-1}(x^{2012})) - k\times (\int \frac{x^{2012} \tan^{-1}(x^{2012})}{1+x^{4024}} dx)\] In order to match with the given formula, the value of k should be \(\frac{1}{2012}\). Therefore, the correct answer is (a) \([1/(2012)]\).

Key concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are essential tools in calculus, particularly when dealing with integrals and derivatives. These functions are the inverses of the usual trigonometric functions like sine, cosine, and tangent. In operations involving calculus, the inverse tangent function, denoted as \(\tan^{-1}(x)\), is especially popular due to its straightforward derivative and integral properties.

The derivative of \(\tan^{-1}(x)\) is given by:

• \[\frac{d}{dx}(\tan^{-1}(x)) = \frac{1}{1+x^2}\]

This formula is invaluable when integrating functions that transform into the form \(\frac{1}{1+y^2}\), as it directly leads to the result \(\tan^{-1}(y) + C\), where \(C\) is the integration constant.

Inverse trigonometric functions, like \(\tan^{-1}(x^{2012})\), often appear in complex calculus problems because the integration of their derivatives results in easily recognizable forms. Such features make them advantageous in simplifying integrations that might otherwise be more convoluted.

Integration Techniques

Integration techniques are crucial when solving calculus problems involving complex functions, such as those with polynomial powers and inverse trigonometric functions. Among these techniques, integration by parts stands out as a versatile method used to break down the product of two functions.

The integration by parts formula is:

• \[\int u \, dv = uv - \int v \, du\]

In the provided solution, we assign \(u\) to be \(\tan^{-1}(x^{2012})\). This choice leverages the straightforward differentiation of the inverse tangent function. Meanwhile, \(dv\) was chosen to be \(\frac{x^{2012}}{1+x^{4024}}dx\), simplifying the original integral into more manageable parts.

This technique also involves transforming the variable through substitution. Recognizing patterns such as \((1 + x^{2012})^2\) can streamline our approach, allowing us to solve parts of the integral we might otherwise ignore or find complicated. Integration by parts essentially "gives back" some control over how we simplify difficult integrals.

Differentiation

Differentiation is a foundational concept in calculus, focusing on how a function changes at any given point. It involves calculating the derivative of a function, which represents the instantaneous rate of change. This is crucial when using techniques like integration by parts, where derivatives help identify components like \(du\).

For the function \(\tan^{-1}(x^{2012})\), its derivative, \(du\), becomes:

• \[du = \frac{2012x^{2011}}{1+x^{4024}}dx\]

Understanding the derivative allows us to accurately apply integration techniques like integration by parts. The derivative here signifies how \(\tan^{-1}(x^{2012})\) changes with respect to \(x\), providing insight into the behavior of the function.

Differentiation is often the first step in breaking down a complex calculus problem into easier parts. It serves as a bridge between the original function and its integral form, ensuring that every component of the sum or product is accounted for. Mastering differentiation is key to tackling a wide range of calculus challenges effectively.