The integral [ (2x - 1) (2x - 1)^2 + 5 4x^2 - 4x + 6 dx] is equal to (where c is a constan…

Question

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)

Accepted Answer

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

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

$$2(2x - 1)2dx = 2t\,dt$$

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

So, $$\int {{{\sqrt {{t^2} - 5} \cos t} \over {2\sqrt {{t^2} - 5} }}dt = {1 \over 2}\sin t + c} $$

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

Solutions · 1