# Help with Trigonometric Integrals

• everestwitman
In summary: Thank you again.In summary, the first problem involved using the substitution u = csc(x) to ultimately solve for the integral of 6 csc^3(x) cot(x) dx. The second problem involved finding the length of a curve using the formula L = ∫sqrt(1 + (d(ln(csc x))^2) dx from x = π/4 to π/2.
everestwitman
Could someone help me with these two problems? I've been at them for an hour, but have very little clue how to go about solving either of them.

## Homework Statement

1)∫ 6 csc^3 (x) cot x dx

## The Attempt at a Solution

6 ∫ csc^3 (x) dx) / tan x

csc^3 / tan x = csc^3 cot x

cot^2 x = csc^2 x - 1
csc^2 x = cot^2 x + 1

csc x cot x (cot^2 x + 1)

u = csc x
du = - csc x cot x dx

## Homework Statement

2)Find the length of the curve: y = ln(csc x), π/4 <= x <= π/2

## The Attempt at a Solution

L = ∫sqrt(1 + (d(ln(csc x))^2) dx from x = π/4 to π/2

d(ln(csc x)/dx = -cot x
(1 + cot x^2) = csc^2

L = ∫ csc x dx from x = π/4 to π/2

everestwitman said:
Could someone help me with these two problems? I've been at them for an hour, but have very little clue how to go about solving either of them.

## Homework Statement

1)∫ 6 csc^3 (x) cot x dx

## The Attempt at a Solution

6 ∫ csc^3 (x) dx) / tan x

csc^3 / tan x = csc^3 cot x

cot^2 x = csc^2 x - 1
csc^2 x = cot^2 x + 1

csc x cot x (cot^2 x + 1)

u = csc x
du = - csc x cot x dx
...

The two problems are not all that closely related.

It's generally better to post these in separate threads.

For the first problem.

The substitution u = csc(x) is a good choice.

As you noted, du = - csc(x) cot(x) dx .

Solve that for dx and plug that into the integral. Then you'll easily see what's left over to complete the substitution.

Thanks. With using your suggestion for a substitution and playing around with the other problem, i have them both figured out.

## 1. What is a trigonometric integral?

A trigonometric integral is an integral that involves trigonometric functions, such as sine, cosine, tangent, etc. It can also involve other algebraic functions and constants.

## 2. Why are trigonometric integrals important?

Trigonometric integrals are important because they allow us to solve problems in physics, engineering, and other fields that involve periodic functions. They also have many applications in calculus and advanced mathematics.

## 3. How do I solve a trigonometric integral?

There is no one-size-fits-all method for solving trigonometric integrals. However, there are some common techniques, such as using trigonometric identities, substitution, or integration by parts. It is important to have a strong understanding of trigonometric functions and algebraic manipulation to successfully solve these integrals.

## 4. What are some tips for solving difficult trigonometric integrals?

Some tips for solving difficult trigonometric integrals include practicing and becoming familiar with trigonometric identities, using substitution to simplify the integral, and breaking down the integral into smaller, more manageable parts. It is also helpful to have a good understanding of the properties of trigonometric functions and their graphs.

## 5. Are there any common mistakes to avoid when solving trigonometric integrals?

Yes, some common mistakes to avoid when solving trigonometric integrals include forgetting to use the chain rule when integrating trigonometric functions, making algebraic mistakes when simplifying the integral, and forgetting to include the constant of integration. It is also important to double-check your work and make sure your final answer is in the correct form.

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