Question

In the following exercises, use appropriate substitutions to express the trigonometric integrals in terms of compositions with logarithms. $$ \int x \csc \left(x^{2}\right) d x $$

Short answer

The integral is \(-\frac{1}{2} \ln |\csc(x^2) + \cot(x^2)| + C\) using substitution.

Step-by-step solution

Recognizing Suitable Substitution

To solve the integral \( \int x \csc(x^2) \, dx \), notice the expression \( x^2 \) inside the cosecant function. This suggests using substitution with \( u = x^2 \).

Find Derivative of Substitution

Calculate the derivative of \( u \). Since \( u = x^2 \), then \( \frac{du}{dx} = 2x \). As such, \( du = 2x \, dx \), or equivalently, \( x \, dx = \frac{1}{2} du \).

Substitute and Simplify the Integral

Substitute \( u = x^2 \) and \( x \, dx = \frac{1}{2} du \) into the integral: \( \int x \csc(x^2) \, dx = \frac{1}{2} \int \csc(u) \, du \). This integral is now simpler to evaluate.

Evaluate the Integral of Cosecant

The integral \( \int \csc(u) \, du \) is a standard integral that evaluates to \( -\ln |\csc(u) + \cot(u)| + C \). Substitute back \( u = x^2 \) to express the integral in terms of \( x \).

Resubstitute u with x

Replace \( u \) with \( x^2 \) to get \( -\frac{1}{2} \ln |\csc(x^2) + \cot(x^2)| + C \) as the solution of the original integral.

Key concepts

Substitution Method

The substitution method is a powerful tool in calculus used to simplify complex integrals. It's akin to unraveling a puzzle by converting it into a more recognizable form. When dealing with trigonometric integrals, substitution can convert a challenging expression into something manageable.

Let's examine how the substitution works in the exercise. The original integral\[ \int x \csc\left(x^2\right) \, dx \]features the expression \(x^2\) within the cosecant function. To make our lives easier, we identify an appropriate substitution: \(u = x^2\). This is because the derivative of \(x^2\) plays nicely into transforming \(x \, dx\), which is required by the structure of the integral.

After setting our substitution, we determine \(du = 2x \, dx\), which we can rearrange to yield \(x \, dx = \frac{1}{2} du\). This transformation simplifies our original integral into\[ \frac{1}{2} \int \csc(u) \, du \]This form is much easier to evaluate, highlighting the effectiveness of substitution in trigonometric integrals.

Cosecant Function

The cosecant function, denoted by \(\csc(x)\), is one of the six fundamental trigonometric functions. It is the reciprocal of the sine function. Mathematically, it is represented as\[ \csc(x) = \frac{1}{\sin(x)} \]This function arises often in integrals involving trigonometric expressions. When integrating the cosecant function, as in\[ \int \csc(u) \, du \]we leverage a known solution for its antiderivative. This solution connects us to logarithmic functions, specifically given by\[ \int \csc(u) \, du = -\ln |\csc(u) + \cot(u)| + C \]Here, \(\cot(u)\) is the cotangent function, another fundamental trigonometric function defined as \(\frac{\cos(u)}{\sin(u)}\). Understanding the relationship between these functions is crucial since it simplifies the integration process and leads directly to the result.

Logarithms

Logarithms are mathematical concepts pivotal in simplifying multiplicative processes into additive ones, thus making complex calculations more manageable. In calculus, they often appear as part of the outcomes of integrals involving certain trigonometric functions.

When we integrated \( \csc(u) \, du \), we obtained:\[ -\ln |\csc(u) + \cot(u)| + C \]This result illustrates the connection between trigonometric functions and logarithms, providing a neat exponential form to express the solution. Logarithms help relate the multiplicative interplay of \(\csc(u)\) and \(\cot(u)\) in a simplified logarithmic expression.

In the context of our problem, once we substituted back \(u = x^2\), the logarithmic expression elegantly summarized our solution, encapsulating the integral's original complexity in a comprehensible form. The use of logarithms in integration, especially with trigonometric functions, provides an essential bridge to solve difficult integrals effectively.