Question

Evaluate the integral
(tan x + cot x)^2

Answer 1

(tan x+1/tan x)^2dx
{(tan^2+1)/tanx)}^2dx
now put tan x=t
sec^2x dx=dt
or,(1+tan^2x)dx=dt
or, dx=dt/(1+t^2).
put these value in equation
{(t^2+1)/t}^2.dt/(1+t^2)
or (t^2+1)/t^2 dt
or, 1/t^2 dt +1dt
or,-1/t+t+c
or -1/tanx +tan x+c
or , tanx-cotx+c

Answer 2

verification....
diff.of tanx-cotx+c=(secx)^2+(cosecx)^2
=1+(tanx)^2+1+(cotx)^2
=(tanx)^2+(cotx)^2+2 tanx cotx
=(tan x + cot x)^2

Answer 3

convert tanx +cotx to \[sinx/cosx+cosx/sinx = (\sin ^{2}x + \cos ^{2}x)/(cosxsinx)= 2/\sin2x\]
Now take the integral of \[4/\sin ^{2}2x\] because sinxcosx =sin2x/2

Answer 4

forget all that