Question

Evaluate the following integrals : (i) \(\int \frac{d \theta}{2-3 \cos 2 \theta}\) (ii) \(\int \frac{(1-\sin \theta)}{2 \cos \theta+3} \mathrm{~d} \theta\)(iii) \(\int \frac{\cos x d x}{5-3 \cos x}\) (iv) \(\int \frac{\cos x}{\cos 2 x} d x\)

Short answer

For each of the following integrals, identify the integration technique used and write down its final result: (i) \(\int \frac{d \theta}{2 - 3 \cos 2 \theta}\) (ii) \(\int \frac{(1-\sin\theta)}{2\cos\theta+3} \,d \theta \) (iii) \(\int \frac{\cos x d x}{5-3 \cos x}\) (iv) \(\int \frac{\cos x}{\cos 2 x} d x\) The integration techniques used for the integrals are as follows: (i) Double angle identity and substitution Final result: \(-\frac{1}{\sqrt{6}}\sin^{-1}\left(\frac{\sqrt{6}}{\sqrt{5}}\cos\theta\right) + C\) (ii) Trigonometric identity and substitution Final result: \(\theta - \sin ^{-1}\left(\frac{3}{\sqrt{13}}\sin \theta\right) + C\) (iii) Substitution Final result: \(\frac{1}{3}\ln|5-3\cos x| + C\) (iv) Double angle identity and substitution Final result: \(-\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\sqrt{2}\cos x}{\sqrt{2}\cos x+1}\right) + C\)

Step-by-step solution

Use the double angle identity

Replace \(\cos 2\theta\) with the double angle identity, \(\cos 2\theta = 2\cos^2\theta - 1\): $$\int \frac{d \theta}{2 - 3(2\cos^2\theta - 1)}$$

Simplify the integrand

Simplify the integrand: $$\int \frac{d \theta}{2 - 6\cos^2\theta + 3} = \int \frac{d \theta}{5 - 6\cos^2\theta}$$

Use the substitution

Use the substitution \(u = \cos\theta\) and \(du = -\sin\theta d\theta\): $$\int \frac{-1}{5 - 6u^2} du$$

Evaluate the integral

Apply partial fraction decomposition, and then find the antiderivative: $$-\frac{1}{\sqrt{6}}\int \frac{1}{\sqrt{1-\frac{u^2}{5/6}}} du = -\frac{1}{\sqrt{6}}\sin^{-1}\left(\frac{\sqrt{6}}{\sqrt{5}}u\right) + C$$

Substitute back

Substitute back \(u = \cos\theta\): $$-\frac{1}{\sqrt{6}}\sin^{-1}\left(\frac{\sqrt{6}}{\sqrt{5}}\cos\theta\right) + C$$ (ii) \(\int \frac{(1-\sin \theta)}{2 \cos \theta+3} \mathrm{~d} \theta\)

Use trigonometric identity and substitution

Use the Pythagorean identity \(\cos^2\theta = 1 - \sin^2\theta\) and then apply the substitution \(u = \sin\theta\) and \(du = \cos\theta d\theta\): $$\int \frac{1 - u}{2\sqrt{1-u^2}+3}du$$

Evaluate the integral

This integral can be evaluated using the method of integration by substitution and then finding the antiderivative. The final result is: $$ \theta - \sin ^{-1}\left(\frac{3}{\sqrt{13}}\sin \theta\right) + C$$ (iii) \(\int \frac{\cos x d x}{5-3 \cos x}\)

Use substitution

Apply the substitution \(u = 3 \cos x\) and \(du = -3\sin x dx\): $$-\frac{1}{3}\int \frac{\sin x}{5-u} du$$

Evaluate the integral

Evaluate the integral: $$-\frac{1}{3}(-\ln|5-u|) + C = \frac{1}{3}\ln|5-3\cos x| + C$$ (iv) \(\int \frac{\cos x}{\cos 2 x} d x\)

Use the double angle identity

Replace \(\cos 2x\) with the double angle identity, \(\cos 2x = 2\cos^2 x - 1\): $$\int \frac{\cos x}{2\cos^2 x - 1} d x$$

Use substitution

Apply the substitution \(u = \cos x\) and \(du = -\sin x dx\): $$-\int \frac{1}{2u^2 - 1} du$$

Evaluate the integral

Evaluate the integral: $$-\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\sqrt{2}u}{\sqrt{2}u+1}\right) + C$$

Substitute back

Substitute back \(u = \cos x\): $$-\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\sqrt{2}\cos x}{\sqrt{2}\cos x+1}\right) + C$$

Key concepts

Integration by Substitution

Integration by substitution, also known as u-substitution, is a method used in integral calculus to simplify the process of finding antiderivatives. This technique transforms a complex integral into a simpler one by introducing a new variable. To apply integration by substitution effectively, you need to identify a part of the integrand that can be substituted with a new variable, typically denoted as u. Then, you derive a differential equation for du, representing the differential part of the original variable.

Let's analyze an example from the exercise provided: (iii) \(\int \frac{\cos x \, dx}{5-3 \cos x}\) Here, the substitution \(u = 3 \cos x\) is applied, making the integral more approachable. With \(du = -3\sin x \, dx\), the integrand changes in terms of u, and thus, simplifies the integration process.

As the substitution is reversed after integration, this method not only aids in calculation but also ensures that the result remains in terms of the original variable. This is particularly beneficial for definite integrals, where the limits of integration must be converted to the new variable u as well.

Trigonometric Identities

Trigonometric identities are formulas involving trigonometric functions which allow us to simplify expressions and perform calculations that would otherwise be difficult or impossible. There are numerous identities, but some of the most commonly used in calculus include Pythagorean identities, angle sum and difference formulas, double angle formulas, and half-angle formulas.

Looking at the solutions provided, we see a use of a double angle identity: (iv) \(\int \frac{\cos x}{\cos 2 x} \, dx\)The identity used here is \(\cos 2x = 2\cos^2 x - 1\), which simplifies the integrand. Knowing these identities is crucial when dealing with integrals that involve trigonometric functions because you can often transform the integrand into a more manageable form for integration.

Furthermore, the way trigonometric identities are creatively used in calculus to modify expressions is a testament to their power and shows why it is important for students to have a strong grasp of them.

Partial Fraction Decomposition

Partial fraction decomposition is a technique used to break down complex rational expressions into simpler fractions, which can often be integrated more easily. This method is particularly useful when dealing with integrals of rational functions, where the numerator and the denominator are polynomials.

In the initial exercise (i), we see that after using substitution, the integral takes the form of a rational function:(i) \(-\frac{1}{\sqrt{6}}\int \frac{1}{\sqrt{1-\frac{u^2}{5/6}}} \, du\)The process of partial fraction decomposition would first involve expressing the integrand as a sum of simpler fractions. After decomposing the fractions, each term can usually be integrated using standard techniques.

This method is powerful and often unveils simpler integration paths that were not initially apparent. It is key to remember that partial fraction decomposition requires the degree of the numerator to be less than the degree of the denominator; otherwise, polynomial long division must be performed first.