Math, asked by ruchirajani1110, 7 months ago

integrate (2tanx- 3 cotx)² dx = ?​

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Answered by viajaypkawle67
1

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 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 = ∫tan² x dx + ∫cot² x - ∫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 - ∫dx + ∫(cosec² x dx - ∫dx - 2 ∫ dx

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

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

∫sec² x. dx = tan x + c1, ∫(cosec² x dx = -cot x + c2 and ∫ dx = x + c3

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)

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