Asked by Wawen
                Please solve it in step by step.
integrate: ∫(cot2x-csc2x)^2 dx
            
        integrate: ∫(cot2x-csc2x)^2 dx
Answers
                    Answered by
            Steve
            
    ∫(cot2x-csc2x)^2 dx
∫ cot^2(2x) - 2cot(2x)csc(2x) + csc^2(2x) dx
∫ 2csc^2(2x) - 1 - 2cos(2x)/sin^2(2x) dx
Now everything is standard forms, so you get
-cot(2x) - x + 1/sin(2x)
csc(2x) - cot(2x) - x
or, you can go on to get
(1-cos(2x))/sin(2x) - x
(1 - 2cos^2(x) + 1)/(2sinx cosx) - x
(2 - 2cos^2(x))/(2sinx cosx) - x
sinx/cosx - x
tanx - x
+ C!
    
∫ cot^2(2x) - 2cot(2x)csc(2x) + csc^2(2x) dx
∫ 2csc^2(2x) - 1 - 2cos(2x)/sin^2(2x) dx
Now everything is standard forms, so you get
-cot(2x) - x + 1/sin(2x)
csc(2x) - cot(2x) - x
or, you can go on to get
(1-cos(2x))/sin(2x) - x
(1 - 2cos^2(x) + 1)/(2sinx cosx) - x
(2 - 2cos^2(x))/(2sinx cosx) - x
sinx/cosx - x
tanx - x
+ C!
                    Answered by
            Bosnian
            
    OR
cot ( 2x ) - csc ( 2x ) = cos ( 2x ) / sin ( 2x ) - 1 / sin ( 2x ) =
[ cos ( 2x ) - 1 ) ] / sin ( 2x )
Since the:
cos ( 2x ) = cos ^ 2 x - sin ^ 2 x
sin ( 2 x ) = 2 sin x cos x
you can write:
cot ( 2x ) - csc ( 2x ) = ( cos ^ 2 x - sin ^ 2 x - 1 ) / 2 sin x cos x =
[ ( - 1 + cos ^ 2 x ) - sin ^ 2 x ) ] / 2 sin x cos x =
[ - ( 1 - cos ^ 2 x ) - sin ^ 2 x ) ] / 2 sin x cos x =
( - sin ^ 2 x - sin ^ 2 x ) / 2 sin x cos x =
- 2 sin ^ 2 x / 2 sin x cos x =
- 2 sin x * sin x / 2 sin x cos x =
- sin x / cos x = - tan x
So:
cot ( 2x ) - csc ( 2x ) = - tan x
Now:
[ cot ( 2x ) - csc ( 2x ) ] ^ 2 = ( - tan x ) ^ 2
[ cot ( 2x ) - csc ( 2x ) ] ^ 2 = tan ^ 2 x
ç [ cot ( 2x ) - csc ( 2x ) ] ^ 2 dx = ç tan ^ 2 x dx =
ç ( sin ^ 2 x / cos ^ 2 x ) dx =
ç [ ( 1 - cos ^ 2 x ) / cos ^ 2 x ] dx =
ç [ ( 1 / cos ^ 2 x - cos ^ 2 x / cos ^ 2 x ) ] dx =
ç ( sec ^ 2 x - 1 ) dx =
ç sec ^ 2 x dx - ç 1 * dx =
ç sec ^ 2 x dx - ç dx =
tan x - x + C
    
cot ( 2x ) - csc ( 2x ) = cos ( 2x ) / sin ( 2x ) - 1 / sin ( 2x ) =
[ cos ( 2x ) - 1 ) ] / sin ( 2x )
Since the:
cos ( 2x ) = cos ^ 2 x - sin ^ 2 x
sin ( 2 x ) = 2 sin x cos x
you can write:
cot ( 2x ) - csc ( 2x ) = ( cos ^ 2 x - sin ^ 2 x - 1 ) / 2 sin x cos x =
[ ( - 1 + cos ^ 2 x ) - sin ^ 2 x ) ] / 2 sin x cos x =
[ - ( 1 - cos ^ 2 x ) - sin ^ 2 x ) ] / 2 sin x cos x =
( - sin ^ 2 x - sin ^ 2 x ) / 2 sin x cos x =
- 2 sin ^ 2 x / 2 sin x cos x =
- 2 sin x * sin x / 2 sin x cos x =
- sin x / cos x = - tan x
So:
cot ( 2x ) - csc ( 2x ) = - tan x
Now:
[ cot ( 2x ) - csc ( 2x ) ] ^ 2 = ( - tan x ) ^ 2
[ cot ( 2x ) - csc ( 2x ) ] ^ 2 = tan ^ 2 x
ç [ cot ( 2x ) - csc ( 2x ) ] ^ 2 dx = ç tan ^ 2 x dx =
ç ( sin ^ 2 x / cos ^ 2 x ) dx =
ç [ ( 1 - cos ^ 2 x ) / cos ^ 2 x ] dx =
ç [ ( 1 / cos ^ 2 x - cos ^ 2 x / cos ^ 2 x ) ] dx =
ç ( sec ^ 2 x - 1 ) dx =
ç sec ^ 2 x dx - ç 1 * dx =
ç sec ^ 2 x dx - ç dx =
tan x - x + C
                    Answered by
            Steve
            
    I like it!
But the half-angle formula would have gone directly to
(cos2x - 1)/sin2x = -tanx
Then we'd have
∫tan^2(x) dx
= ∫sec^2(x)-1 dx
= tanx - x + C
    
But the half-angle formula would have gone directly to
(cos2x - 1)/sin2x = -tanx
Then we'd have
∫tan^2(x) dx
= ∫sec^2(x)-1 dx
= tanx - x + C
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