Question

Evaluate the following integrals. $$\int \cos ^{-1} x d x$$

Short answer

Question: Evaluate the integral of the inverse cosine function with respect to x. Answer: The integral of the inverse cosine function with respect to x is: $$\int \cos^{-1} x dx = x\cos^{-1}x + 2\sqrt{1 - x^2} + C$$ Where C is the constant of integration.

Step-by-step solution

Identify the integration by parts formula

Integration by parts formula is given by: $$\int u dv = uv - \int v du$$ We will use this formula to evaluate the given integral.

Define u and dv

Choose the two parts for integration by parts: $$u = \cos^{-1} x$$ $$dv = dx$$

Calculate du

Find the derivative of \(\cos^{-1} x\) with respect to x: $$x = \cos(\cos^{-1} x)$$ Implicitly differentiate both sides with respect to x: $$1 = \frac{-1}{\sqrt{1 - x^2}} dx$$ Thus, the differential is: $$du = \frac{-1}{\sqrt{1 - x^2}} dx$$

Calculate v

Integrate to find v: $$v = \int dx = x$$

Apply the integration by parts formula

Now we have all the necessary components to apply the integration by parts formula: $$\int \cos^{-1} x dx = \int u dv = uv - \int v du$$ Substitute the values of u, dv, du, and v into the formula and simplify: $$\int \cos^{-1} x dx = x\cos^{-1}x - \int x\left(\frac{-1}{\sqrt{1 - x^2}} dx\right)$$

Evaluate the remaining integral

Evaluate the remaining integral: $$\int \frac{-x}{\sqrt{1 - x^2}} dx$$ Let \(s = 1 - x^2\), then \(-2x dx = ds\). Thus, the integral becomes: $$\int \frac{1}{\sqrt{s}} \frac{1}{2} ds$$ This is a standard integral: $$\int \frac{1}{\sqrt{s}} ds = 2\sqrt{s} + C$$ Substitute back \(s = 1 - x^2\): $$2\sqrt{1 - x^2} + C$$

Combine the results

Combine the results from Step 5 and Step 6: $$\int \cos^{-1} x dx = x\cos^{-1}x + 2\sqrt{1 - x^2} + C$$ Where C is the constant of integration.

Key concepts

Definite Integrals

Definite integrals calculate the net area under a curve between two points.They require upper and lower limits, giving a specific numerical value as the result.Unlike indefinite integrals, which include a constant of integration, definite integrals provide a precise outcome.
For example, if you wanted to calculate the area under a curve from point A to point B, you would use definite integrals.The formula generally has the following form:

• \( \int_{a}^{b} f(x) \, dx \)

Here, \(a\) and \(b\) are the limits of integration. These limits are used to determine the specific interval over which you want to find the area.The problems in calculus often involve calculating definite integrals to solve real-world issues, like finding distances or areas under curves.Although the exercise above was an indefinite integral, understanding definite integrals is essential because it forms the foundation for many advanced topics in calculus.

Trigonometric Integration

Trigonometric integration refers to the techniques used to integrate functions involving trigonometric expressions.These methods are crucial because trigonometric functions appear often in calculus problems, especially in physics and engineering.
In the exercise, we dealt with the inverse cosine function, \( \cos^{-1}(x) \), demonstrating an important aspect of trigonometric integration.The integration of trigonometric functions usually involves identities and substitutions.For instance:

• If faced with \( \int \sin(x) \cos(x) \, dx \), you might use identities to simplify.
• Substituting with known values or other trigonometric identities can also make integration easier.

Trigonometric integration is foundational for more complex calculus problems, and mastering it requires practice with different identities and techniques.Understanding and applying these techniques helps simplify and solve design, wave, and circular motion problems in various fields.

Integration Techniques

Integration techniques include a range of methods used to solve integrals, making it one of the most versatile topics in calculus.The exercise presented the method of integration by parts, a valuable technique for integrals involving products of functions.
Integration by parts is based on the formula:

• \( \int u \, dv = uv - \int v \, du \)

This approach is particularly useful when the integral isn't straightforward, allowing the problem to be broken down into simpler parts.In the step-by-step solution, \( u \) was chosen as \( \cos^{-1}(x) \), making the derivative and substitution a part of solving the integral.Other popular integration techniques include:

• Substitution: Useful for trigonometric or exponential functions, transforming the integral into a simpler form.
• Partial Fractions: Breaks down complex rational expressions for integration.
• Trigonometric Identities: Simplifies integrals by using identities to reduce complexity.

Understanding various integration techniques enables a student to tackle a broad array of integrals.These methods are essential in solving both theoretical math problems and practical applications across many disciplines.