Help with Trigonometric Integrals

[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

Homework Equations

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

Homework Equations

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

 

[SammyS]

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

Homework Equations

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.

 

[everestwitman]

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