Easy looks integral. Help

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In summary, the power of the csc function is even, so when solving for du, use the u-substitution u = cot(x).
  • #1
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Hello I have this integral and I can't solve it


[tex]\int\csc^6 x dx[/tex]


Well i first start with trig identnity [tex]\csc x = \frac{1}{sen x}[/tex] so my first attempt to slove it looks like this

[tex]\int\frac{dx}{(\sin^2 x)^3}[/tex]

then i subsituted with [tex]\sin^2 x = \frac{1 - \cos 2x}{2}[/tex]

and i get

[tex]8\int\frac{dx}{(1-cos 2x)^3} [/tex] and i saw that i was going to

nowhere so i try another way

i took the function and use the trig identity to get this

[tex]\int\csc^6 x dx[/tex] =[tex]\int \frac{\sec^6 x}{\tan^6 x} dx [/tex]

[tex]u = \tan x[/tex] [tex]du = \sec^2 x[/tex]

then

[tex]\int\frac{du^3}{tan^6 x}[/tex]

and i get stucked again. any ideas?
 
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  • #2
alba_ei said:
i took the function and use the trig identity to get this

[tex]\int\csc^6 x dx[/tex] =[tex]\int \frac{\sec^6 x}{\tan^6 x} dx [/tex]

That looks most promising.

If [tex]u = \tan{x}[/tex], then [tex]du = \sec^2{x}dx[/tex] (Don't forget the dx)

Now, rewrite the integral as

[tex]\int{\frac{\sec^4{x}\sec^2{x}dx}{\tan^6{x}}}[/tex]

Now, think of an identity that relates [tex]\sec{x}[/tex] and [tex]\tan{x}[/tex].
 
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  • #3
Make [itex] \tan x =t [/itex] and then u'll get 3 simple integrals.
 
  • #4
alba_ei said:
Hello I have this integral and I can't solve it[tex]\int\csc^6 x dx[/tex]
Here's the trick, if the power of csc function is even, you can use the u-substitution u = cot(x).
And if the power of sec function is even, you can use the u-substitution u = tan(x)
So, in your problem, the power of csc function is even, we use the u-substitution: u = cot(x)
du = -csc2(x) dx, so your integral will become:
[tex]\int \csc ^ 6 (x) dx = \int ( \csc ^ 4 (x) \times \csc ^ 2 (x) ) dx = \int {( \csc ^ 2 (x) )} ^ 2 \times \csc ^ 2 (x) dx = \int {\left( 1 + \cot ^ 2 (x) \right)} ^ 2 \times \csc ^ 2 (x) dx[/tex]
[tex]= -\int {(1 + u ^ 2)} ^ 2 du = ...[/tex]
Can you go from here? :)
 

1. What is an easy looks integral?

An easy looks integral is a type of mathematical integral that can be solved quickly and easily using specific techniques and formulas. These integrals typically involve simple functions and have a straightforward approach to solving them.

2. Why is it important to know how to solve an easy looks integral?

Understanding how to solve an easy looks integral is important because it allows you to quickly find the area under a curve and solve many real-world problems involving rates of change, such as velocity and acceleration.

3. What are some common techniques for solving an easy looks integral?

Some common techniques for solving an easy looks integral include using basic integration rules, substitution, and integration by parts. It is also helpful to have a good understanding of basic algebra, trigonometry, and calculus concepts.

4. Can an easy looks integral be solved using software or calculators?

Yes, many software programs and calculators have built-in functions that can solve easy looks integrals. However, it is still important to understand the underlying concepts and techniques used in solving these integrals.

5. What are some tips for solving an easy looks integral efficiently?

Some helpful tips for solving an easy looks integral efficiently include practicing regularly, breaking down the integral into smaller parts, and using known formulas and techniques. It is also important to carefully check your work and make sure to simplify your final answer.

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