If αβγ∫sinxsin3x+cos3xdx=αloge|1+tanx|+βloge|1-tanx+tan2x|+γtan-1(2tanx-13)+C, when C is constant of integration, then the value of 18(α + β + γ2) is ______. - Mathematics (JEE Main)

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If `int sinx/(sin^3x + cos^3x)dx = α log_e |1 + tan x| + β log_e |1 - tan x + tan^2x| + γ tan^-1 ((2tanx - 1)/sqrt(3)) + C`, when C is constant of integration, then the value of 18(α + β + γ2) is ______.

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Solution

If `int sinx/(sin^3x + cos^3x)dx = α log_e |1 + tan x| + β log_e |1 - tan x + tan^2x| + γ tan^-1 ((2tanx - 1)/sqrt(3)) + C`, when C is constant of integration, then the value of 18(α + β + γ2) is 3.

Explanation:

I = `int (sinx/cos^3x)/(1 + tan^3x)dx`

= `int (tanx.sec^2x)/((tanx + 1)(1 + tan^2x - tanx))dx`

Let tan x = t

`\implies` sec3x.dx = dt

= `int t/((t + 1)(t^2 - t + 1))dt`

Let `t/((t + 1)(t^2 - t + 1)) = A/(t + 1) + (B(2t - 1))/(t^2 - t + 1) + C/(t^2- t + 1)`

`\implies` A(t2 – t + 1) + B(2t – 1)(t + 1) + C(t + 1) = t

`\implies` t2(A + 2B) + t(–A + B + C) + A – B + C = t

∴ A + 2B = 0  ...(i)

–A + B + C = 1  ...(ii)

A – B + C = 0  ...(iii)

`\implies` C = `1/2`

`\implies` A – B = `-1/2`  ...(iv)

Solving equations (i) and (iv), we get

B = `1/6`, A = `-1/3`

∴ I = `-1/3int(dt)/(1 + t) + 1/6int(2t - 1)/(t^2 - t + 1)dt + 1/2int (dt)/(t^2 - t + 1)`

= `-1/3log_e|(1 + tan x)| + 1/6 log_e|tan^2x - tan x + 1| + 1/2. 2/sqrt(3) tan^-1(((tan x - 1/2))/(sqrt(3)/2))`

= `-1/3log_e|(1 + tan x)| + 1/6log_e|tan^2x - tan x + 1| + 1/sqrt(3)tan^-1 ((2tanx - 1)/sqrt(3)) + C`

α = `-1/3`, β = `1/6`, γ = `1/sqrt(3)`

So, 18(α + β + γ2) = `18(-1/3 + 1/6 + 1/3)` = 3

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