- #1

- 39

- 1

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?

[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?

Last edited: