Question

Evaluate the following integrals : (i) \(\int \frac{\cos x}{\sqrt{1+\cos x}} d x\) (ii) \(\int \frac{\mathrm{dx}}{\sin \mathrm{x} \sin (\mathrm{x}+\alpha)}\) (iii) \(\int\\{1+\cot (x-\alpha) \cot (x+\alpha\\} d x\)

Short answer

Question: Evaluate the following integrals: (i) \(\int \frac{\cos x}{\sqrt{1+\cos x}} dx\) (ii) \(\int \frac{dx}{\sin(x+\alpha)}\) (iii) \(\int\{1+\cot(x-\alpha)\cot(x+\alpha)\} dx\) Answer: (i) \(\int \frac{\cos x}{\sqrt{1+\cos x}} dx = -\frac{2}{3}((1+\cos x)^{\frac{3}{2}}-(1+\cos x)^{\frac{1}{2}})+C\) (ii) \(\int \frac{dx}{\sin(x+\alpha)} = \frac{1}{2}(1 - \cos\alpha) \ln\left|\frac{\sin x + \sqrt{\sin^2 x - \cos^2 x \sin^2 \alpha}}{\sin x - \sqrt{\sin^2 x - \cos^2 x \sin^2 \alpha}}\right| + C\) (iii) \(\int\{1+\cot(x-\alpha)\cot(x+\alpha)\} dx = x + C\)

Step-by-step solution

Identify the substitution

Observe that the denominator of the fraction is \(\sqrt{1+\cos x}\). We can use the substitution \(u = 1 + \cos x\), which simplifies the expression.

Differentiate the substitution

We must differentiate the substitution \(u = 1 + \cos x\) with respect to x: \(\frac{du}{dx} = -\sin x\).

Change the integral in terms of u

Using the substitution, we have: \(u = 1 + \cos x \Rightarrow \cos x = u - 1\) \(\frac{du}{dx} = -\sin x \Rightarrow dx = -\frac{du}{\sin x}\) Now, rewrite the integral: \(\int \frac{\cos x}{\sqrt{1+\cos x}} dx = \int \frac{u-1}{\sqrt{u}} (-\frac{du}{\sin x})\)

Simplify the integral

We can remove the \(\sin x\) term in the denominator, as it will be canceled out: \(\int \frac{u-1}{\sqrt{u}} (-\frac{du}{\sin x}) = -\int \frac{u-1}{\sqrt{u}} du\)

Integrate and make the reverse substitution

Integrate the expression and then substitute back in terms of x: \(-\int \frac{u-1}{\sqrt{u}} du = -\frac{2}{3}(u^{\frac{3}{2}}-u^{\frac{1}{2}})+C = -\frac{2}{3}((1+\cos x)^{\frac{3}{2}}-(1+\cos x)^{\frac{1}{2}})+C\) So, the final solution for the integral in part (i) is: \(\int \frac{\cos x}{\sqrt{1+\cos x}} d x = -\frac{2}{3}((1+\cos x)^{\frac{3}{2}}-(1+\cos x)^{\frac{1}{2}})+C\) #Part (ii)#

Apply the sine addition formula

First, apply the sine addition formula to the denominator: \(\sin(x+\alpha) = \sin x \cos\alpha + \cos x \sin \alpha\)

Simplify the integral

Rewrite the integral with the sine addition formula: \(\int \frac{dx}{\sin x \cdot (\sin x \cos\alpha + \cos x \sin \alpha)}\)

Use the substitution trick

Multiply the numerator and the denominator of the integrand by \(\sin x - \cos x \sin \alpha\): \(\int \frac{(\sin x - \cos x \sin \alpha)dx}{\sin^2 x - \cos^2 x \sin^2 \alpha}\)

Integrate by partial fractions

Decompose the integrand into partial fractions and integrate: \(\int \frac{(1 - \cos\alpha)\: dx}{\sin^2 x - \cos^2 x \sin^2 \alpha} = \frac{1}{2}(1 - \cos\alpha) \ln\left|\frac{\sin x + \sqrt{\sin^2 x - \cos^2 x \sin^2 \alpha}}{\sin x - \sqrt{\sin^2 x - \cos^2 x \sin^2 \alpha}}\right| + C\) #Part (iii)#

Rewrite the cotangent functions using sum-to-product formula

Rewrite the cotangent functions using the sum-to-product formula: \(\cot(x-\alpha) = \frac{\cos x \cos \alpha + \sin x \sin \alpha}{\sin x \cos \alpha - \cos x \sin \alpha}\) \(\cot(x+\alpha) = \frac{\cos x \cos \alpha - \sin x \sin \alpha}{\sin x \cos \alpha + \cos x\sin\alpha}\)

Simplify the expression inside the integral

Multiply the cotangent functions and simplify the expression: \((1+\cot(x-\alpha)\cot(x+\alpha))\) \(=\left(1+\frac{(\cos x\cos\alpha+\sin x\sin\alpha)^2-\sin^2 x\sin^2\alpha}{(\sin x\cos\alpha-\cos x\sin\alpha)^2}\right)\) \(=\frac{(\sin x\cos\alpha-\cos x\sin\alpha)^2}{(\sin x\cos\alpha-\cos x\sin\alpha)^2}\)

Integrate the simplified expression

Now we can integrate the simplified expression: \(\int\{1+\cot(x-\alpha)\cot(x+\alpha)\} dx\) $=\int\frac{(\sin x\cos\alpha-\cos x\sin\alpha)^2}{(\sin x\cos\alpha-\cos x\sin\alpha)^2} dx \\ =\int dx$ So, the final solution for the integral in part (iii) is: \(\int\\{1+\cot (x-\alpha) \cot (x+\alpha\\} d x=x+C\)

Key concepts

Integration Techniques

Integration is a fundamental concept in calculus that represents the process of finding the area under or above a curve and, more broadly, solving for the antiderivative of a function. An array of integration techniques is used to tackle integrals that are not immediately solvable by a basic formula. Among such techniques are substitution, integration by parts, partial fractions, and the use of trigonometric identities. Becoming proficient in these methods is essential for students looking to solve more complex integrals, as exemplified by the exercise.

Two of the aforementioned techniques — trigonometric integrals and the substitution method — feature in the textbook exercise. In solving these problems, we face the challenge of transforming an integral into a more manageable form. By employing a substitution, we can simplify the integrand or even convert the integral into a standard form, which is easier to integrate.

Trigonometric Integrals

Integrals involving trigonometric functions, such as sine, cosine, and tangent, require special techniques to solve. These trigonometric integrals can often be simplified using trigonometric identities such as the Pythagorean identity, double-angle formulas, and sum-to-product formulas. For example, in the given exercise for part (iii), the sum-to-product formulas are used to rewrite the expression involving cotangent functions before integration.

When faced with trigonometric expressions that do not readily yield to integration, one can manipulate these expressions to obtain an equivalent expression that is easier to integrate. This manipulation often involves factoring, using known identities, or applying a substitution that relates to the specific trigonometric function. Understanding the nuances of these identities and their application is key to solving trigonometric integrals efficiently.

Substitution Method

The substitution method, also known as u-substitution, is a powerful tool for simplifying integrals. It involves replacing a portion of the integral with a new variable to simplify the integrand into a form that's easier to integrate. This method can be thought of as the reverse chain rule of differentiation.

In Step 1 of our exercise, we identify a suitable substitution, such as \(u = 1 + \text{cos} x\), which simplifies the denominator. We then differentiate our substitution to express \(dx\) in terms of \(du\), as done in Step 2. Next, we perform the substitution and rewrite our integral solely in terms of \(u\), which often reveals a simpler antiderivative to integrate. Once integrated, we substitute back the original expression for \(u\) to arrive at the final answer.

Students should always be attentive to the possibility of a substitution which can convert a complex integral into a recognizable form, making it a powerful fixture in their mathematical toolkit.