# Evaluate the indefinite integral of y=2sin x- 2tan^2x?

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Since the integral is additive, we'll get:

Int [2sin x- 2(tan x)^2]dx = Int 2sin x dx - Int 2(tan x)^2 dx (*)

We'll solve the first integral from the right side:

Int 2sin dx = 2Int sin x dx= -2 cos x + C (1)

Int 2(tan x)^2 dx = 2Int [(sec x)^2 - 1]dx

2Int [(sec x)^2 - 1]dx = 2Int (sec x)^2 dx - 2Int dx

2Int [(sec x)^2 - 1]dx = 2 tan x - 2x + C (2)

We'll substitute (1) and (2) in (*):

Int [2sin x- 2(tan x)^2]dx = -2 cos x- 2 tan x + 2x + C

Int [2sin x- 2(tan x)^2]dx = 2(x - tan x - cos x) + C

**The indefinite integral of the given function 2sin x- 2(tan x)^2 is Int [2sin x- 2(tan x)^2]dx = 2(x - tan x - cos x) + C.**