Question

integral of (tanx+cotx)²​

Answer 1

Answer:

Step-by-step explanation:

Let I = ∫(tan x - cot x)² . dx

= ∫(tan² x + cot² x - 2 tanx . cotx) . dx [Using the formula (a-b)² = a² + b² - 2ab]

According to a theorem on indefinite integration, integral of the sum of functions is sum of integrals of individual functions. Further, we have the trigonometric relation, cot x = 1/tan x , that is tan x. cot x = 1. We then have,

I = ∫tan²x  dx + ∫cot² x dx - ∫2. 1. dx                                          

= ∫(sec² x -1) dx + ∫(cosec² x -1) dx - 2 ∫ dx

                                                [∵tan² x = sec² x -1, cot² x = cosec² x -1]

= ∫sec² x. dx - ∫1 dx + ∫(cosec² x dx - ∫1 dx - 2 ∫1  dx

= ∫sec² x. dx + ∫(cosec² x dx - 4 ∫1 dx …………………………………………………..…….(1)

Now each of the above integral is of standard form and we have,

where c1, c2, and c3 are arbitrary constants. Substituting these values in (1),

I = tan x + c1 -cot x + c2 - 4x + c3 = tan x - cot x - 4x + c

Hence, ∫(tan x - cot x)² . dx = tan x - cot x - 4x + c (Answer)