Question

Evaluate the following integrals: (i) \(\int \frac{\mathrm{d} \theta}{1-\sin ^{4} \theta}\) (ii) \(\int \frac{d x}{\sin ^{4} x+\cos ^{4} x}\) (iii) \(\int \frac{d x}{1-\cos ^{4} x}\) (iv) \(\int \frac{d \theta}{\left(\operatorname{acos}^{2} \theta+b \sin ^{2} \theta\right)^{2}}\)

Short answer

Question: Evaluate the following definite integral: $$\int \frac{\mathrm{d} \theta}{1-\sin^4 \theta}$$ Answer: $$-\frac{1}{2}\left( \sin^{-1}(2\cos^2 \theta-1) - \frac{1}{2} \ln\left|\frac{2\cos^2 \theta-1}{\sin^2 \theta}\right| + c \right)$$

Step-by-step solution

Simplify the expression

We start by factoring the denominator. We use the identity \(\sin^2 \theta + \cos^2 \theta = 1\) and write the integral as: $$\int \frac{\mathrm{d} \theta}{1-\sin^4 \theta} = \int \frac{\mathrm{d} \theta}{1-(1-\cos^2 \theta)^2}$$

Further simplify the expression

We can now expand the denominator and simplify the expression as follows: $$\int \frac{\mathrm{d} \theta}{1-(1-\cos^2 \theta)^2} = \int \frac{\mathrm{d} \theta}{(2\cos^2 \theta-1)(1-\cos^2 \theta)}$$

Use substitution

Let's make a substitution, \(t = \cos^2 \theta\). Then, we have \(\dfrac{dt}{d\theta} = -2 \cos \theta \sin \theta\) or \(d\theta = \frac{dt}{-2 \cos \theta \sin \theta}\), and the integral becomes: $$\int \frac{1}{(2t-1)(1-t)} \cdot \frac{dt}{-2 \sqrt{t(1-t)}}$$

Simplify the expression

By simplifying the resulting expression, we get: $$-\frac{1}{2} \int \frac{dt}{\sqrt{t(1-t)}(2t-1)(1-t)}$$

Partial fraction decomposition

Next, we perform partial fraction decomposition to rewrite the integral as the sum of simpler integrals: $$-\frac{1}{2} \int \frac{A}{\sqrt{t(1-t)}} + \frac{B}{2t-1} + \frac{C}{1-t} dt$$ After solving for constants A, B, and C, the integral becomes: $$-\frac{1}{2} \int \frac{1}{\sqrt{t(1-t)}} - \frac{1}{2(2t-1)} + \frac{1}{2(1-t)} dt$$

Evaluate the integral

Now, we evaluate each of these three simpler integrals separately, using standard integrals and substitution methods: $$-\frac{1}{2}\left( \int \frac{1}{\sqrt{t(1-t)}} dt - \int \frac{1}{2(2t-1)} dt + \int \frac{1}{2(1-t)} dt \right)$$ The result is: $$-\frac{1}{2}\left( \sin^{-1}(2t-1) - \frac{1}{2} \ln\left|\frac{2t-1}{1-t}\right| + c \right)$$

Replace t with the original variable

Finally, we substitute back \(\cos^2 \theta\) for t: $$-\frac{1}{2}\left( \sin^{-1}(2\cos^2 \theta-1) - \frac{1}{2} \ln\left|\frac{2\cos^2 \theta-1}{\sin^2 \theta}\right| + c \right)$$

Key concepts

Integral Calculus

Integral calculus is a fundamental branch of mathematics that deals with finding the antiderivatives of functions. This process, known as integration, is essential for solving problems involving area, volume, displacement, and many other concepts that arise in physics and engineering.

For example, in the exercise provided, integral calculus is used to evaluate expressions with complex trigonometric functions. The process begins by simplifying the integrand, the function inside the integral, using trigonometric identities and algebraic manipulation. Simplification makes the integration process more straightforward, often converting a difficult integral into a series of standard integrals that can be evaluated more easily.

The integral calculus technique showcased in the step by step solution also includes the use of substitutions which can transform the variable of integration, simplifying the integrand. After the substitution, the problem is solved by integrating simpler expressions and then substituting back to the original variable.

Trigonometric Substitution

Trigonometric substitution is a technique used to evaluate integrals that involve radicals or certain trigonometric expressions. In this method, a variable is replaced with a trigonometric function that simplifies the integrand, leveraging the properties of triangles and trigonometric identities.

In the exercise at hand, trigonometric substitution could be applied to the integral by setting a trigonometric function equal to a part of the integrand. For instance, if a segment of the integrand resembles part of a Pythagorean identity, you can use trigonometric substitution to replace that segment, simplifying the integral's evaluation.

Example of Trigonometric Substitution

Considering the identity \(\theta^2 = 1 - \text{cos}^2 \theta\), we can substitute \(t = \text{cos}^2 \theta\) in the integration process. This transforms the integral into a function of \(t\), which is often easier to integrate. After integrating with respect to the new variable \(t\), we then replace \(t\) with the original trigonometric expression to complete the solution.

Partial Fraction Decomposition

Partial fraction decomposition is an algebraic method used to break down complex rational functions into the sum of simpler fractions. This makes the task of integration more manageable when facing complex rational expressions.

In the provided example, partial fraction decomposition is employed once the integrand has been transformed by substitution and simplification. By expressing the integrand as a sum of simpler fractions, each with its own constant in the numerator, the integral is then divided into the sum of simpler integrals. These are much easier to evaluate using standard integration techniques.

Breaking Down the Integrand

After finding the constants \(A, B, C\) for each of the fractions, you are left with individual integrals that are standardized. Each fraction represents a known or simpler integral form which is then integrated separately. The complexity of the original integral is greatly reduced by this decomposition, making the solution accessible.

Standard Integrals

Standard integrals refer to a collection of integrals of basic functions that are commonly encountered in calculus. These integrals have known solutions and serve as a reference for solving more complex integrals. They are essential to solving integration problems efficiently and are often used in conjunction with other techniques like substitution and partial fraction decomposition.

In the steps provided for the exercise solution, after implementing partial fraction decomposition, standard integrals are used to evaluate each term of the decomposed function. The solutions to these standard integrals are often straightforward and may involve basic antiderivatives, natural logarithms, and inverse trigonometric functions.

Utilizing Standard Integrals

Leveraging standard integrals, the solution to the exercise further simplifies and becomes a direct application of these known results. For instance, integrals resulting in inverse trigonometric functions such as \(\text{sin}^{-1}(x)\) or natural logarithms like \(\text{ln}(x)\) are quickly recognized and solved. This highlights the power of knowing and applying standard integrals in the integration process.