# What is the indefinite integral of the function 3sinx-4tan^2x ?

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### 2 Answers

We have to find the integral of f(x) = 3*sin x - 4*(tan x)^2

f(x) = 3*sin x - 4*(tan x)^2

=> 3*sin x - 4*[(sec x)^2 - 1]

Int [ f(x) dx] = Int [3*sin x - 4*[(sec x)^2 - 1]

=> 3*Int [ sin x dx] - 4*Int [ (sec x)^2 dx] + 4* Int [ 1 dx]

=> 3*(- cos x) - 4* tan x + 4*x + C

**The required integral is -3*(cos x) - 4* tan x + 4*x + C**

We'll apply the property of integral to be additive:

Int [3sin x- 4(tan x)^2]dx = Int 3sin x dx - Int 4(tan x)^2 dx (*)

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

Int 3sin dx = 3Int sin x dx= -3 cos x + C (1)

Int 4(tan x)^2 dx = 4Int [(sec x)^2 - 1]dx

4Int [(sec x)^2 - 1]dx = 4Int (sec x)^2 dx - 4Int dx

4Int [(sec x)^2 - 1]dx = 4 tan x - 4x + C (2)

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

Int [3sin x- 4(tan x)^2]dx = -3 cos x- 4 tan x + 4x + C

**The anti-derivative of the trigonometric function 3sin x- 4(tan x)^2 is Int [3sin x- 4(tan x)^2]dx = -3 cos x- 4 tan x + 4x + C.**