Question

Evaluate the following integrals: \(\mathrm{dx}\) (i) \(\int \frac{\sin x+\cos x}{\sqrt[3]{\sin x-\cos x}}\) (ii) \(\int e^{x} \tan \left(e^{x}\right) \sec \left(e^{x}\right) d x\) (iii) \(\int\left(3 \sin x \cos ^{2} x-\sin ^{3} x\right) d x\) (iv) \(\int \frac{\sec x \operatorname{cosec} x}{\log (\tan x)} d x\)

Short answer

Question: Evaluate the following integrals: (i) \(\int \frac{\sin x + \cos x}{\sin x - \cos x}\, dx\) (ii) \(\int e^x \tan e^x \sec e^x \, dx\) (iii) \(\int (3 \sin x \cos^2 x - \sin^3 x) \, dx\) (iv) \(\int \frac{\sec x \csc x}{\log (\tan x)} \, dx\) Answer: (i) \(\frac{3}{2}(\sin x - \cos x)^\frac{2}{3} + C\) (ii) \(\sec(e^{x}) + C\) (iii) \(-\frac{4}{3}\cos^3 x + \cos x + C\) (iv) \(-\log(\tan x) \cot x + \ln|\sin x| + C\)

Step-by-step solution

(i) Simplification

Since the given integral consists of trigonometric functions, we can simplify the expression using some trigonometric identities. Let's start with substitution, let \(u = \sin x - \cos x\). Therefore, \(du= (\cos x + \sin x) \, dx\). Now the integral becomes: \(\int \frac{du}{\sqrt[3]{u}}\)

(i) Evaluation

The simplified integral is easy to evaluate, we just need to apply power rule: \(\int \frac{1}{\sqrt[3]{u}} du= \int u^{-\frac{1}{3}} du = \frac{3}{2}u^{\frac{2}{3}} + C = \frac{3}{2}(\sin x - \cos x)^\frac{2}{3} + C\)

(ii) Substitution

The given integral consists of exponential and trigonometric functions. To evaluate such integrals, we can use substitution method. Let \(u = e^x\), so \(du = e^x\, dx\). Now the integral becomes: \(\int \tan u \sec u \, du\)

(ii) Evaluation

The simplified integral can now be easily evaluated using the substitution method. We can simply integrate the tangent and secant functions: \(\int \sec u \tan u \, du = \sec u + C\) Substitute back \(u = e^x\): \(\sec(e^{x}) + C\)

(iii) Simplification

The given integral consists of trigonometric functions. We can simplify the expression by factoring out \(\sin x\): \(\int(3 \sin x \cos^2 x - \sin^3 x) dx = \int \sin x (3\cos^2 x - \sin^2 x) dx\) We can use the identity \(\sin^2 x + \cos^2 x = 1\), so \(\sin^2 x = 1 - \cos^2 x\). Substitute this in the above integral: \(\int \sin x (3\cos^2 x - (1 - \cos^2 x)) dx = \int \sin x (4\cos^2 x - 1) dx\)

(iii) Evaluation

Now we can use substitution method to evaluate the simplified integral. Let \(u = \cos x\), so \(du = -\sin x\, dx\). Now the integral becomes: \(-\int (4u^2 - 1) du\) Evaluate the integral: \(-\int (4u^2 - 1) du = -\left(\frac{4}{3}u^3 - u\right) + C\) Substitute back \(u = \cos x\): \(-\frac{4}{3}\cos^3 x + \cos x + C\)

(iv) Substitution

The given integral has secant and cosecant functions in the numerator. To evaluate such integrals, we can use substitution method. Let \(u = \log(\tan x)\), so \(du = \frac{1}{\tan x}\sec^2 x\, dx\) Now rewrite the given integral in terms of the substitution: \(\int \frac{\sec x \csc x}{u} \times \frac{u \tan x}{\sec^2 x} dx = \int \csc x u \, du\)

(iv) Evaluation

Now we can use integration by parts to evaluate this integral. Let \(v = u\) so \(dv = du\), and let \(dw = \csc x du\), so \(w = -\cot x\). Apply integration by parts: \(\int u dw = uw - \int v dw = -u\cot x - \int -\cot x du = -u\cot x + \int \cot x du\) Now, integrate the remaining integral: \(\int \cot x du = \int \frac{\cos x}{\sin x} du = \ln|\sin x| + C\) So, the final answer is: \(-u\cot x + \ln|\sin x| + C = -\log(\tan x) \cot x + \ln|\sin x| + C\)

Key concepts

Trigonometric Integrals

When solving integrals involving trigonometric functions, recognizing patterns and identities can greatly simplify the process. Trigonometric integrals typically contain sin, cos, tan, sec, csc, and cot functions. An effective approach for integrating these functions is to use trigonometric identities to rewrite the integral in a more manageable form.

For instance, in the exercise, the integral \(\int(3 \sin x \cos^2 x - \sin^3 x) dx\) benefits from simplification by factoring out \(\sin x\) and employing the Pythagorean identity \(\sin^2 x = 1 - \cos^2 x\). This leads to an integral that can be handled using substitution or other integration methods, demonstrating how simplification can facilitate the solution of trigonometric integrals.

Substitution Method

The substitution method is a powerful tool for evaluating integrals, especially when dealing with complex or compound functions. By changing variables, we can transform a difficult integral into a simpler one. The general strategy is to substitute a part of the integrand with a new variable \(u\) so that the differential \(du\) corresponds to another part of the integrand.

This technique was applied in evaluating the integral \(\int \frac{\sin x+\cos x}{\sqrt[3]{\sin x-\cos x}} dx\), where \(u = \sin x - \cos x\) simplified the expression greatly, allowing the use of a straightforward power rule.

Integration by Parts

Integration by parts is another essential method of integral calculus, particularly useful when the integrand is a product of two functions whose integration in their current form is not straightforward. It is based on the product rule for differentiation and can be represented as \(\int u dv = uv - \int v du\), where \(u\) and \(dv\) are parts of the integrand chosen strategically.

The final integral in the exercise, \(\int \frac{\sec x \operatorname{cosec} x}{\log (\tan x)} d x\), exemplifies the usage of integration by parts. By recognizing \(u = \log(\tan x)\) and \(dv = \csc x\), the integral is manageable and can be simplified to reach a solution. This example shows integration by parts breaking down a complex integral into more elementary parts for easier computation.