Question

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

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

Answer 1

The correct option is B $\frac{1}{2}\sin \sqrt{(2x-1{)}^2+5}+c$

$\int \frac{(2x-1)\cos \sqrt{(2x-1{)}^2+5}}{\sqrt{4x^2-4x+6}}dx$
Put $(2x-1{)}^2+5=t^2$
$2(2x-1)dx=2tdt$
$\Rightarrow \int \frac{\cos t}{t}\times \frac{t}{2}dx=\frac{1}{2}\sin t+c$
$=\frac{1}{2}\sin \sqrt{(2x-1{)}^2+5}+c$