The integral ∫ ( 2x - 1 ) cos ( 2x - 1 ) ^ 2 + 5 4 x ^ 2 - 4x + 6 dx is equal to (where c is a constant of integration)

Correct answer: A. (1)/(2)sin ( 2x - 1 ) ^ 2 + 5 + c ∫ ( 2x - 1 ) cos ( 2x - 1 ) ^ 2 + 5 ( 2x - 1 ) ^ 2 + 5 dx ( 2x - 1 ) ^ 2 + 5 = t ^ 2 2(2x - 1)2dx = 2tdt 2 t ^ 2 - 5 dx = tdt So ∫ t ^ 2 - 5 t 2 t ^ 2 - 5 dt = (1)/(2) t + c = (1)/(2)sin ( 2x - 1 ) ^ 2 + 5 + c

JEE Main 202118 MarchMorning Shift

Exam: JEE Main
Year: 2021
Date: 18 March
Shift: Morning Shift
Subject: Mathematics
Chapter: Indefinite Integration

JEE Main 2021 · 18 March · Morning Shift · Mathematics · Indefinite Integration

The integral ∫ ( 2x - 1 ) cos ( 2x - 1 ) ^ 2 + 5 4 x ^ 2 - 4x + 6...

The integral

∫ \dfrac{( 2x - 1 ) cos\sqrt{( 2x - 1{)}^{2} + 5}}{\sqrt{4{x}^{2} - 4x + 6}}dx

is equal to (where

c

is a constant of integration)

Options

A $\dfrac{1}{2}sin\sqrt{( 2x - 1{)}^{2} + 5} + c$
B $\dfrac{1}{2}cos\sqrt{( 2x + 1{)}^{2} + 5} + c$
C $\dfrac{1}{2}cos\sqrt{( 2x - 1{)}^{2} + 5} + c$
D $\dfrac{1}{2}sin\sqrt{( 2x + 1{)}^{2} + 5} + c$

Correct answer

A. $\dfrac{1}{2}sin\sqrt{( 2x - 1{)}^{2} + 5} + c$

Step-by-step solution

$∫ \dfrac{( 2x - 1 ) cos\sqrt{( 2x - 1{)}^{2} + 5}}{\sqrt{( 2x - 1{)}^{2} + 5}}dx$

$( 2x - 1{)}^{2} + 5 = {t}^{2}$

$2\left(2x - 1\right)2dx = 2tdt$

$2\sqrt{{t}^{2} - 5}dx = tdt$

So $∫ \dfrac{\sqrt{{t}^{2} - 5}\cos t}{2\sqrt{{t}^{2} - 5}}dt = \dfrac{1}{2}\sin t + c$

$= \dfrac{1}{2}sin\sqrt{( 2x - 1{)}^{2} + 5} + c$

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