Question

int dx sqrt 1-cos 4x​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{\displaystyle \int\,\sqrt{1-cos(4x)}\,dx}$

$\sf{=\displaystyle \int\,\sqrt{2\,sin^2(2x)}\,\,dx}$

$\sf{=\displaystyle \int\,\sqrt{2}\cdot\,sin(2x)\,\,dx}$

$\sf{=\sqrt{2}\cdot\displaystyle \int\,sin(2x)\,\,dx}$

$\sf{=\sqrt{2} \cdot\dfrac{-cos(2x)}{2}+C}$

$\sf{=-\dfrac{1}{\sqrt{2}}\,\, cos(2x)+C}$