The integral

Question:

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 $\mathrm{c}$ is a constant of integration)

 

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

  2. $\frac{1}{2} \cos \sqrt{(2 x+1)^{2}+5}+c$

  3. $\frac{1}{2} \cos \sqrt{(2 x-1)^{2}+5}+c$

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


Correct Option: 1

Solution:

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

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

$2(2 \mathrm{x}-1) 2 \mathrm{dx}=2 \mathrm{t} \mathrm{dt}$

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

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

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

 

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