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

Correct answer: (A) 1/2sin√((2x-1)^2+5)+c. ∫((2x-1)cos√((2x-1)^2+5)√((2x-1)^2+5)dx(2x-1)^2+5=t^222x-1 2dx=2 tdt2√(t^2-5)dx=tdtSo ∫√(t^2-5)cost)/(2√(t^2-5))dt=1/2sint+c =1/2sin√((2x-1)^2+5)+c

Q35 / 90 · JEE Main 2021 (18 Mar Shift 1) · Mathematics

The integral $\int \frac{(2 x - 1) \cos \sqrt{(2 x - 1)^{2} + 5}}{\sqrt{4 x^{2} - 4 x + 6}} d x$ is equal to (where $c$ is a constant of integration)

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

Solution

$\int \frac{(2 x - 1) \cos \sqrt{(2 x - 1)^{2} + 5}}{\sqrt{(2 x - 1)^{2} + 5}} d x$

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

$2 (2 x - 1) 2 d x = 2 t d t$

$2 \sqrt{t^{2} - 5} d x = t d t$

So $\int \frac{\sqrt{t^{2} - 5} \cos t}{2 \sqrt{t^{2} - 5}} d t = \frac{1}{2} \sin t + c$

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

Hence, the correct answer is option: 'A'.

The integral ∫(2x-1)cos(2x-1)2+54x2-4x+6dx is equal to (where c is a constant of integration)

Solution

Relevant Questions

The integral $\int \frac{(2 x - 1) \cos \sqrt{(2 x - 1)^{2} + 5}}{\sqrt{4 x^{2} - 4 x + 6}} d x$ is equal to (where $c$ is a constant of integration)