Question

Evaluate the following integrals. $$\int \csc ^{10} x \cot x d x$$

Short answer

Answer: \(\csc ^2x = 1 + \cot ^2x\)

Step-by-step solution

Rewrite exponent

Rewrite the integrand as a product of \(\csc ^2x\) and \(\csc ^8x\): $$\int \csc ^{10} x \cot x d x = \int \csc ^2 x \cot x \csc ^8 x d x$$

Use substitution

Substitute \(\csc ^2x\) with \(1+ \cot ^2x\): $$\int \csc ^{10} x \cot x d x = \int (1 + \cot ^2x) \cot x \csc ^8x d x$$

Split the integral

Split the integral into two parts: $$\int \csc ^{10} x \cot x d x = \int \cot x \csc ^8x d x + \int \cot^3 x \csc ^8 x d x$$

Integration by parts

Perform integration by parts on the first integral. Let \(u = \cot x\) and \(d v = \csc ^8 x d x\). Then, \(du = -\csc ^2 x d x\), \(v = -\csc ^7 x / 7\). So we have: $$\int \cot x \csc ^8x d x = -\frac{1}{7} \cot x \csc ^7x + \frac{1}{7} \int \csc ^2x \csc ^7 d x$$

Apply substitution again

Substitute \(\csc ^2x = 1 + \cot ^2 x\) in the remaining integral: $$\int \csc ^2 x \csc ^7x d x = \int (\csc ^7 x + \cot^2 x \csc ^7 x) d x$$

Split the integral

Split the integral into two parts: $$\int \csc ^2 x \csc ^7x d x = \int \csc ^7 x d x + \int \cot^2 x \csc ^7 x d x$$ Now, we need to find the integral of the new expressions.

Simplify and find antiderivative

Simplify and find the antiderivative of the new expressions: $$\int \csc ^7 x d x$$ $$\int \cot^2 x \csc ^7 x d x$$ Now let's go back and put everything together.

Combine the results

Combine the results of the calculations in Steps 4-7: $$\int \csc ^{10} x \cot x d x = -\frac{1}{7} \cot x \csc ^7x + \frac{1}{7} \int \csc ^2x \csc ^7 d x$$ $$\int \csc ^{10} x \cot x d x = -\frac{1}{7} \cot x \csc ^7x + \frac{1}{7} (\int \csc ^7 x d x + \int \cot^2 x \csc ^7 x d x)$$ By finding the antiderivatives and substituting back, we will have the full solution to the given integral.

Key concepts

Trigonometric Integrals

Trigonometric integrals are integrals that involve trigonometric functions such as sin, cos, tan, sec, csc, and cot. In our exercise, we are dealing with the integral \( \int \csc^{10} x \cot x \, dx \). The challenge here is to simplify and integrate expressions with higher powers of these trigonometric functions.

One useful technique when dealing with trigonometric integrals is rewriting the expressions in terms of other trigonometric functions, like sec and tan for \( \csc \) and \( \cot \) respectively, or breaking them into smaller, more manageable parts. Our initial step is to rewrite \( \csc^{10} x \cot x \) in terms of \( \csc^2 x \) and \( \csc^8 x \), facilitating later steps. This splitting allows us to isolate simpler integrals that can be tackled step by step using other integration techniques like substitution or integration by parts.

Integration by Parts

Integration by parts is a powerful technique based on the product rule for differentiation. It's expressed by the formula:\[\int u \, dv = uv - \int v \, du\]In our problem, we apply integration by parts specifically to the integral \( \int \cot x \, \csc^8 x \, dx \). This is a classic situation where there's a product of functions - one that becomes simpler when differentiated and another that we can integrate easily.

Here, we set \( u = \cot x \) which gives \( du = -\csc^2 x \, dx \), and \( dv = \csc^8 x \, dx \) where we find \( v = -\frac{1}{7} \csc^7 x \). This choice cleverly reduces the degree of the trigonometric power, breaking the integral into manageable parts. By systematically applying integration by parts, we're simplifying heavy calculus problems into smaller, bite-sized calculations.

Substitution Method

The substitution method simplifies integrals by making suitable variable substitutions. Our integral problem makes use of smart substitutions, especially those that harness trigonometric identities. For the integral \( \int \csc^{10} x \cot x \, dx \), a substitution of \( \csc^2 x = 1 + \cot^2 x \) plays a crucial role.

This substitution helps in expressing the integral in terms of \( \cot x \), entering a recognizable form absent the overbearing powers originally present. The substitution allows us to break the integration task into smaller, more manageable pieces by dealing with simpler forms. Whether it’s breaking down the integral or changing the variable to delve into easier derivatives, substitution is the translator between complicated integrals and their simpler counterparts.

Antiderivatives

An antiderivative of a function is the opposite of a derivative, signifying the function that "undoes" differentiation. The process of finding an antiderivative is called integration. In the given exercise, the task boils down to finding the antiderivative of expressions after breaking them using integration by parts and substitution.

The final step involves combining the individually integrated or simplified parts into a complete solution. For the function \( \int \csc^7 x \, dx \) and \( \int \cot^2 x \, \csc^7 x \, dx \), finding antiderivatives often requires recognizing patterns or leveraging known integral forms. Once these antiderivatives are found and combined, the solution of the original integral is complete. Antiderivatives are foundational in evaluating definite and indefinite integrals, reflecting the core summation concept over continuous functions.