Question

Please solve it in step by step.
integrate: ∫(cot2x-csc2x)^2 dx

Answer 1

∫(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!

Answer 2

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

Answer 3

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