If ∫ x ^ 3 x + ^ 3 x dx = α _ e | 1 + x | + β _ e |1 - x + ^ 2 x| + γ ^ - 1 ( 2 x - 1 3 ) + C, when C is constant of integration, then the value of 18(α + β + γ ^ 2 ) is

Correct answer: 0 ∫ x ^ 3 x + ^ 3 x dx = α _ e | 1 + x | + β _ e |1 - x + ^ 2 x| + γ ^ - 1 ( 2 x - 1 3 ) + C, . . . . . . . . . . (i) Let, I = ∫ x ^ 3 x + ^ 3 x dx I = ∫ x sec ^ 2 x ^ 3 + 1 dx Put x = t ⇒ ^ 2 xdx = dt = ∫ tdt t ^ 3 + 1 = ∫ t ( t + 1 ) ( t ^ 2 - t + 1) dt Now t ( t + 1 ) ( t ^ 2 - t + 1) = (A)/(t + 1) + Bt + C t ^ 2 - t + 1 t = A( t ^ 2 - t + 1) + ( Bt + C ) ( t + 1 ) t = t ^ 2 (A + B) - t(A - B - C) + A + C Comparing on both sides A + B = 0, - A + B + C = 1, A + C = 0 Solving these equations A = - (1)/(3), B = (1)/(3), C = (1)/(3) Hence I = ∫ \ (- (1)/(3))/(t + 1) + (1)/(3)( t + 1 t ^ 2 - t + 1 )\ dt = - (1)/(3) ∫ (1)/((t + 1))dt + (1)/(3) ∫ (1)/(2)(2t - 1) + (3)/(2) t ^ 2 - t + 1 dt = - (1)/(3) ∫ (1)/((t + 1))dt + (1)/(6) ∫ 2t - 1 ( t ^ 2 - t + 1) + (1)/(2) ∫ dt (t - (1)/(2)) ^ 2 + ( 3 2 ) ^ 2 = - (1)/(3) |t + 1| + (1)/(6) | t ^ 2 - t + 1| + (1)/(2) · 2 3 ^ - 1 ( 2t - 1 3 ) + C = - (1)/(3) | x + 1| + (1)/(6) | ^ 2 x - x + 1| + 1 3 ^ - 1 ( 2 x - 1 3 ) + C From equation (i) , Comparing from both sides ⇒ α = - (1)/(3), β = (1)/(6) and γ = 1 3 So 18(α + β + γ ^ 2 ) = 18(- (1)/(3) + (1)/(6) + (1)/(3)) = 3

JEE Main 202131 AugustEvening Shift

Exam: JEE Main
Year: 2021
Date: 31 August
Shift: Evening Shift
Subject: Mathematics
Chapter: Indefinite Integration

JEE Main 2021 · 31 August · Evening Shift · Mathematics · Indefinite Integration

If ∫ x ^ 3 x + ^ 3 x dx = α _ e | 1 + x | + β _ e |1 - x + ^ 2 x| +...

∫ \dfrac{\sin x}{\sin^{3}x + \cos^{3}x}dx = \alpha{\log}_{e} | 1 + \tan x | + \beta{\log}_{e}\left|1 - \tan x + \tan^{2}x\right| + \gamma\tan^{- 1}\left(\dfrac{2\tan x - 1}{\sqrt{3}}\right) + C,

when

C

is constant of integration, then the value of

18\left(\alpha + \beta + {\gamma}^{2}\right)

Correct answer

Step-by-step solution

∫ \dfrac{\sin x}{\sin^{3}x + \cos^{3}x}dx = \alpha{\log}_{e} | 1 + \tan x | + \beta{\log}_{e}\left|1 - \tan x + \tan^{2}x\right| + \gamma\tan^{- 1}\left(\dfrac{2\tan x - 1}{\sqrt{3}}\right) + C, . . . . . . . . . . \left(i\right) $Let,$ I = ∫ \dfrac{\sin x}{\sin^{3}x + \cos^{3}x}dx $ $ I = ∫ \dfrac{\tan x{sec}^{2}x}{\tan^{3} + 1}dx $Put$ \tan x = t ⇒ \sec^{2}xdx = dt $$ = ∫ \dfrac{tdt}{{t}^{3} + 1} = ∫ \dfrac{t}{( t + 1 ) \left({t}^{2} - t + 1\right)}dt $ Now$ \dfrac{t}{( t + 1 ) \left({t}^{2} - t + 1\right)} = \dfrac{A}{t + 1} + \dfrac{Bt + C}{{t}^{2} - t + 1} $$ t = A\left({t}^{2} - t + 1\right) + ( Bt + C ) ( t + 1 ) $$ t = {t}^{2}\left(A + B\right) - t\left(A - B - C\right) + A + C $Comparing on both sides$ A + B = 0, - A + B + C = 1, A + C = 0 $Solving these equations $ A = - \dfrac{1}{3}, B = \dfrac{1}{3}, C = \dfrac{1}{3} $ Hence$ I = ∫ \left\{\dfrac{- \dfrac{1}{3}}{t + 1} + \dfrac{1}{3}\left(\dfrac{t + 1}{{t}^{2} - t + 1}\right)\right\}dt $ $ = - \dfrac{1}{3} ∫ \dfrac{1}{\left(t + 1\right)}dt + \dfrac{1}{3} ∫ \dfrac{\dfrac{1}{2}\left(2t - 1\right) + \dfrac{3}{2}}{{t}^{2} - t + 1}dt $$ = - \dfrac{1}{3} ∫ \dfrac{1}{\left(t + 1\right)}dt + \dfrac{1}{6} ∫ \dfrac{2t - 1}{\left({t}^{2} - t + 1\right)} + \dfrac{1}{2} ∫ \dfrac{dt}{{\left(t - \dfrac{1}{2}\right)}^{2} + {\left(\dfrac{\sqrt{3}}{2}\right)}^{2}}$ $= - \dfrac{1}{3}\ln \left|t + 1\right| + \dfrac{1}{6}\ln \left|{t}^{2} - t + 1\right| + \dfrac{1}{2} \cdot \dfrac{2}{\sqrt{3}}\tan^{- 1}\left(\dfrac{2t - 1}{\sqrt{3}}\right) + C$

$= - \dfrac{1}{3}\ln \left|\tan x + 1\right| + \dfrac{1}{6}\ln \left|\tan^{2}x - \tan x + 1\right| + \dfrac{1}{\sqrt{3}}\tan^{- 1}\left(\dfrac{2\tan x - 1}{\sqrt{3}}\right) + C$

From equation \left(i\right) $, Comparing from both sides $ ⇒ \alpha = - \dfrac{1}{3}, \beta = \dfrac{1}{6} $and$ \gamma = \dfrac{1}{\sqrt{3}} $ So$ 18\left(\alpha + \beta + {\gamma}^{2}\right) $$ = 18\left(- \dfrac{1}{3} + \dfrac{1}{6} + \dfrac{1}{3}\right) $$ = 3$

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