Question

vEvaluate the following integrals: (i) \(\int_{0}^{1} \frac{2 x+3}{5 x^{2}+1}\) (ii) \(\int_{-1}^{1} 5 x^{4} \sqrt{x^{5}+1} d x\) (iii) \(\int_{0}^{2 \pi} \frac{\cos x}{\sqrt{4+3 \sin x}} d x\) (iv) \(\int_{0}^{\sqrt[3]{\pi^{2}}} \sqrt{x} \cos ^{2}\left(x^{\frac{3}{2}}\right) d x\)

Short answer

A1: Partial fraction decomposition. Q2: What substitution is used for the second integral? A2: The substitution used is \(u = x^5 + 1\). Q3: What trigonometric identity is applied for the third integral? A3: The trigonometric identity applied is \(\cos^2(u) = \frac{1 + \cos(2u)}{2}\). Q4: What is the value of the third integral? A4: The value of the third integral is zero. Q5: What are the new limits of integration for the fourth integral after the substitution? A5: The new limits of integration are \(u(0) = 0\) and \(u(\sqrt[3]{\pi^2}) = \pi\).

Step-by-step solution

Integral (i) - Evaluate the Simple Rational Function

For the first integral, we have a simple rational function: (i) \(\int_{0}^{1} \frac{2x+3}{5x^2+1} dx\) This can be directly integrated using partial fraction decomposition: (i) \(\int_{0}^{1} \frac{2x+3}{5x^2+1} dx = \int_{0}^{1} \frac{A}{x+B} + \frac{C}{x+D} dx\) To find the values of A, B, C, and D, multiply both sides by \((5x^2+1)\) and equate the numerators. Then, integrate the resulting expression and apply the limits of integration.

Integral (ii) - Evaluate with Substitution

For the second integral, we need to perform a substitution in order to evaluate the integral: (ii) \(\int_{-1}^{1} 5x^4 \sqrt{x^5+1} dx\) Let \(u = x^5 + 1\). Then, \(\frac{du}{dx} = 5x^4\) and \(dx = \frac{du}{5x^4}\). The new limits of integration are \(u(-1) = 0\) and \(u(1) = 2\). The integral becomes: (ii) \(\int_{0}^{2} \sqrt{u} \frac{1}{5} du = \frac{1}{5} \int_{0}^{2} u^{\frac{1}{2}} du\) Now, we will use the power rule for integration and apply the limits of integration.

Integral (iii) - Evaluate with Trigonometric Identity

For the third integral, we need to apply a trigonometric identity before evaluating the integral: (iii) \(\int_{0}^{2 \pi} \frac{\cos x}{\sqrt{4+3 \sin x}} dx\) Let's perform a substitution \(u = \sin x\). Then, \(du = \cos x \, dx\). The new limits of integration are \(u(0) = 0\) and \(u(2 \pi) = 0\). The integral becomes: (iii) \(\int_{0}^{0} \frac{1}{\sqrt{4+3u}} du\) Notice that the limits are the same, so the integral is zero.

Integral (iv) - Evaluate with Substitution

For the fourth integral, we need to perform a substitution to evaluate the integral: (iv) \(\int_{0}^{\sqrt[3]{\pi^2}} \sqrt{x} \cos^2\left(x^{\frac{3}{2}}\right) dx\) Let \(u = x^{\frac{3}{2}}\). Then, \(\frac{du}{dx} = \frac{3}{2}x^{\frac{1}{2}}\) and \(dx = \frac{2}{3}x^{-\frac{1}{2}} du\). The new limits of integration are \(u(0) = 0\) and \(u(\sqrt[3]{\pi^2}) = \pi\). The integral becomes: (iv) \(\int_{0}^{\pi} \cos^2(u) \frac{2}{3} du\) Now, we will use the power reduction formula \(\cos^2(u) = \frac{1 + \cos(2u)}{2}\) to simplify the expression, integrate the resulting expression, and apply the limits of integration.

Key concepts

Partial Fraction Decomposition

Understanding partial fraction decomposition is critical when dealing with rational functions in integral calculus. This method involves breaking down a complex rational expression into simpler fractions that can be integrated more easily. Consider an expression like \( \frac{2x+3}{5x^2+1} \).

First, one must assume that this fraction can be represented as the sum of two or more fractions with simpler denominators. After determining the appropriate form, such as \( \frac{A}{x+B} + \frac{C}{x+D} \), we equate the numerators and solve for the unknown coefficients A, B, C, and D. This reduces the original integral into simpler integrals that can be evaluated using basic integration techniques.

Substitution Method

The substitution method is a powerful tool in integral calculus, especially when dealing with composite functions or functions that involve roots.

For example, when we have an integral like \( \int 5x^4 \sqrt{x^5+1} dx \), we let another variable, typically \( u \), represent the inside of the composite function (\( x^5 + 1 \)) which simplifies the integral. The differentiation of \( u \), \( du = 5x^4 dx\), shows the relationship between \( dx \) and \( du \), allowing for a straightforward substitution. After substituting and finding the new limits of integration, we can often integrate using simpler rules. This method extends beyond simple equations and applies to more complex functions where direct integration is not as apparent.

Trigonometric Identities

Using trigonometric identities simplifies integrals involving trigonometric functions. These identities are relationships between trigonometric functions that allow us to rewrite expressions in different, often simpler, forms.

In exercises such as \( \int \frac{\cos x}{\sqrt{4+3 \sin x}} dx \), substitutions involving these identities can convert the integral into a more manageable one. However, it's significant to be mindful of the limits of integration when using substitution in trigonometric integrals, as they can lead to instances where the integral evaluates to zero, as in this specific example.

Integration Limits

Properly assessing integration limits is essential for correct evaluation of definite integrals. When you apply a substitution, you must adjust the limits accordingly to reflect the change of variable.

For instance, if \( x \) is changed to \( u \) in a derivative, we must find \( u(0) \) and \( u(1) \) to replace the original limits of \( 0 \) and \( 1 \). Incorrect limits can lead to wrong answers. If there's a case with the same upper and lower limits after substitution, like \( \int_{0}^{0} \), we can immediately conclude the integral's value is zero without further calculation.

Power Reduction Formula

The power reduction formula is used to transform powers of sine and cosine into forms that are more conducive to integration. It's a type of trigonometric identity that is particularly useful with even powers.

As an example, \( \cos^2(u) \) can be rewritten as \( \frac{1 + \cos(2u)}{2} \), making the integral much simpler to handle. This is beneficial when dealing with integrals like \( \int_0^\pi \cos^2(u) \, du\) because it converts a trigonometric square into an expression that can be integrated directly using standard techniques. Thus, knowing the power reduction formulas for sine and cosine is a significant asset in solving integral calculus problems.