Question

Evaluate the following integrals: (i) \(\int \frac{\cos 2 x}{(\cos x-\sin x)^{2}}\) (ii) \(\int \frac{d x}{x \sin ^{2}(1+\log x)}\) (iii) \(\int \frac{\mathrm{dx}}{\left(1+\mathrm{x}^{2}\right) \sqrt{\left(\tan ^{-1} \mathrm{x}+3\right)}}\) (iv) \(\int \frac{\left(x^{2}-1\right) d x}{\left(x^{4}+3 x^{2}+1\right) \tan ^{-1}(x+1 / x)}\)

Short answer

Question: Evaluate the following integrals: (i) \(\int \frac{\cos 2 x}{(\cos x-\sin x)^{2}} dx\) (ii) \(\int \frac{d x}{x \sin ^{2}(1+\log x)} dx\) (iii) $\int \frac{\mathrm{dx}}{\left(1+\mathrm{x}^{2}\right) \sqrt{\left(\tan ^{-1} \mathrm{x}+3\right)}} dx$ (iv) $\int \frac{\left(x^{2}-1\right) d x}{\left(x^{4}+3 x^{2}+1\right) \tan ^{-1}(x+1/x)} dx$ Solution: (i) \(-\left(\cos x - \sin x\right) + C\) (ii) \(-\cot (1+\log x) + C\) (iii) \(2\sqrt{\tan^{-1} x + 3} + C\) (iv) Non-elementary (cannot be expressed in terms of elementary functions)

Step-by-step solution

Perform a substitution

Let \(u = \cos x - \sin x\). Then, \(du = (-\sin x - \cos x) dx\). Now we have: \(\int \frac{\cos 2 x}{(\cos x-\sin x)^{2}} dx = \int \frac{-u^2}{u^2} du\)

Integrate

This integral simplifies to \(\int -1 du = -u + C\), where C is the constant of integration.

Undo the substitution

We substitute back u = cos x - sin x, which gives us the final answer: \(-\left(\cos x - \sin x\right) + C\). For (ii):

Perform a substitution

Let \(u = 1 + \log x\). Then, \(du = \frac{1}{x} dx\). Now we have: \(\int \frac{d x}{x \sin^2{u}} = \int \frac{du}{\sin^2{u}}\)

Integrate

This integral is just \(\int \csc^2{u} du\), which has an antiderivative \(-\cot u + C\).

Undo the substitution

We substitute back \(u = 1 + \log x\), which gives us the final answer: \(-\cot (1 + \log x) + C\). For (iii):

Perform a substitution

Let \(t = \tan^{-1} x\). Then, \(dt = \frac{1}{1+x^2} dx\). Now we have: \(\int\frac{dx}{\left(1+x^2\right)\sqrt{\left(\tan^{-1}x+3\right)}} = \int\frac{dt}{\sqrt{t + 3}}\)

Integrate

The integral becomes: \(\int\frac{1}{\sqrt{t + 3}} dt\). This is a standard form with an antiderivative: \(2\sqrt{t + 3} + C\).

Undo the substitution

We substitute back \(t = \tan^{-1} x\), which gives us the final answer: \(2\sqrt{\tan^{-1} x + 3} + C\). For (iv):

Perform a substitution

Let \(t = x+\frac{1}{x}\). Then, \(dt = \left(1-\frac{1}{x^2}\right) dx\). Now we have: $\int \frac{\left(x^{2}-1\right) d x}{\left(x^{4}+3 x^{2}+1\right) \tan ^{-1}(x+1/x)}=\int \frac{(t^2-3) (\left(1-\frac{1}{x^2}\right) dx)}{(t^2+1) \tan ^{-1}(t)}$

Simplify and cancel terms

We notice that \(x^2 - 1 = (x + 1/x)(x - 1/x) = (t)(x - 1/x)\). So we can rewrite the integral as: \(\int\frac{t(x - \frac{1}{x})}{(t^2+1)\tan^{-1}(t)} dt\)

Integrate

We have \(\int\frac{t}{(t^2+1)\tan^{-1}(t)} dt\). This integral is non-elementary, which means that it cannot be expressed in terms of elementary functions. The final answers are: (i) \(-\left(\cos x - \sin x\right) + C\) (ii) \(-\cot (1+\log x) + C\) (iii) \(2\sqrt{\tan^{-1} x + 3} + C\) (iv) Non-elementary (cannot be expressed in terms of elementary functions)

Key concepts

Integration Techniques

Integration, a fundamental concept within calculus, involves finding the antiderivative of a given function. Understanding various integration techniques is crucial for solving complex integrals. Some common techniques include substitution, which simplifies the integrand by changing variables, and integration by parts, which is useful when the integrand is a product of functions. Other methods such as partial fractions are employed when dealing with rational functions. Mastery of these techniques is essential, as they are the building blocks that allow us to tackle more intricate integrals.

In the provided exercises, substitution is used extensively. For instance, in exercise (i), a substitution is made to transform the integral into a simpler form that can be easily integrated. This substitution is powerful as it often converts a complex integral into a basic one, making it a go-to method for many problems.

Trigonometric Substitution

When an integral involves square roots or the squares of variables, trigonometric substitution is an effective technique to simplify the equation. This method uses the properties of trigonometric identities to replace certain expressions with trigonometric functions, which are easier to integrate. The sine and cosine functions, for instance, are particularly useful as their derivatives are easily managed within an integral context.

Exercise (iii) perfectly illustrates the use of trigonometric substitution. By recognizing the term \(1+x^2\) and assigning \(\tan^{-1} x\) to a new variable \(t\), the original complicated expression is reduced to an integral involving \(t\), which we can handle using basic integral rules.

Inverse Trigonometric Functions Integration

Integrating inverse trigonometric functions is another useful skill. These functions, such as \(\tan^{-1} x\), \(\sin^{-1} x\), and \(\cos^{-1} x\), appear frequently in calculus problems, and understanding how to integrate them is vital. The integration of these functions often utilizes their derivatives, which are well-established.

In the exercises (ii) and (iii), inverse trigonometric function integration is showcased. After a substitution, the integrals boil down to forms where the antiderivatives of \(\csc^2 u\) and \(\frac{1}{\sqrt{t + 3}}\) are sought, and recognizing these as derivatives of inverse trigonometric functions simplifies the process significantly.

Non-elementary Integrals

Some integrals are classified as non-elementary because they cannot be expressed in terms of elementary functions, which include polynomials, exponential functions, logarithms, and trigonometric functions. For these challenging integrals, standard techniques may not yield a tidy antiderivative. Instead, we may need to resort to numerical methods or special functions for a solution.

Exercise (iv) presents a non-elementary integral, as indicated in the step-by-step solution. Despite efforts to simplify the integrand via substitution, the resulting integral cannot be expressed using basic elementary functions. Understanding the existence of such non-elementary integrals prepares students for encountering problems that require more advanced methods of integration or approximation techniques.