Integral of (tanx)^4(dx)=A(tanx)^(3)+B(tanx)+f(x) then what is A,B And f(x)???

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... ∫ tan⁴ x dx

= ∫ tan² x. tan² x dx

= ∫ tan² x. ( sec² x - 1 ) dx

= ∫ tan² x. sec² x dx - ∫ tan² x dx

= ∫ tan² x d(tan x) - ∫ ( sec² x - 1 ) dx

= ( tan³ x / 3 ) - ( tan x - x ) + C

= (1/3) tan³ x + (-1) tan x + ( x + C )

= ( A ) tan³ x + ( B ) tan x + ƒ(x) ............ [given]

Hence :

A = 1/3, B = -1, ƒ(x) = x + C .................... Ans.
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Take the derivative of a * tan(x)^3 + B * tan(x) + f(x)

a * 3 * tan(x)^2 * sec(x)^2 + B * sec(x)^2 + f'(x) = tan(x)^4
a * 3 * tan(x)^2 * (1 + tan(x)^2) + B * (1 + tan(x)^2) + f'(x) = tan(x)^4
3a * tan(x)^2 + 3a * tan(x)^4 + B + B * tan(x)^2 + f'(x) = tan(x)^4

3a * tan(x)^4 = tan(x)^4
3a = 1
a = 1/3

3a * tan(x)^2 + B * tan(x)^2 = 0
3a + B = 0
1 + B = 0
B = -1

B + f'(x) = 0
-1 + f'(x) = 0
f'(x) = 1
f(x) = x

int tan^4x dx= A tan^3x +B tanx + f(x)--------------------------(I)
= x- 4/3 tanx + 1/3 tanx ( sec^2x
= x - 4/3 tanx + 1/3 tanx( 1 +tan^2x)
= x - 4/3 tanx + 1/3 tanx + 1/3 tan^3x
= 1/3 tan^3x + tanx(1/3- 4/3) +x ---------------------(II)
comparing (I) & (II)
we have A = 1/3 ,B =( 1/3-4/3) = -1 & f(x) = x answer

tan^4x = tan^4x + tan^2x - tan^2x - 1 +1 = tan^2x (1+ tan^2x) - (tan^2x+1) + 1

integrate ---> A = 1/3 , B = -1 , f(x) = x + C

that one is extremely tricky