JEE MAIN - Mathematics (2021 - 18th March Morning Shift - No. 4)
The integral $$\int {{{(2x - 1)\cos \sqrt {{{(2x - 1)}^2} + 5} } \over {\sqrt {4{x^2} - 4x + 6} }}} dx$$ is equal to (where c is a constant of integration)
$${1 \over 2}\sin \sqrt {{{(2x - 1)}^2} + 5} + c$$
$${1 \over 2}\cos \sqrt {{{(2x + 1)}^2} + 5} + c$$
$${1 \over 2}\cos \sqrt {{{(2x - 1)}^2} + 5} + c$$
$${1 \over 2}\sin \sqrt {{{(2x + 1)}^2} + 5} + c$$
Explanation
$$\int {{{(2x - 1)\cos \sqrt {{{(2x - 1)}^2} + 5} } \over {\sqrt {{{(2x - 1)}^2} + 5} }}} dx$$
$${(2x - 1)^2} + 5 = {t^2}$$
$$2(2x - 1)2dx = 2t\,dt$$
$$2\sqrt {{t^2} - 5} dx = t\,dt$$
So, $$\int {{{\sqrt {{t^2} - 5} \cos t} \over {2\sqrt {{t^2} - 5} }}dt = {1 \over 2}\sin t + c} $$
$$ = {1 \over 2}\sin \sqrt {{{(2x - 1)}^2} + 5} + c$$
$${(2x - 1)^2} + 5 = {t^2}$$
$$2(2x - 1)2dx = 2t\,dt$$
$$2\sqrt {{t^2} - 5} dx = t\,dt$$
So, $$\int {{{\sqrt {{t^2} - 5} \cos t} \over {2\sqrt {{t^2} - 5} }}dt = {1 \over 2}\sin t + c} $$
$$ = {1 \over 2}\sin \sqrt {{{(2x - 1)}^2} + 5} + c$$
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