Question

integration of under √ 1 - cos X​

Answer 1

Solution:

Now, $\mathsf{1-cosx}$

$\quad\mathsf{=1-(1-2\:sin^{2}\dfrac{x}{2})}$

$\quad\mathsf{=1-1+2\:sin^{2}\dfrac{x}{2}}$

$\quad\mathsf{=2\:sin^{2}\dfrac{x}{2}}$

$\implies \mathsf{\sqrt{1-cosx}=\sqrt{2}\:sin\frac{x}{2}}$

$\therefore \mathsf{\int \sqrt{1-cosx}\:dx}$

$\mathsf{=\sqrt{2}\:\int sin\dfrac{x}{2}\:dx}$

$\mathsf{=\sqrt{2}\:\dfrac{1}{\dfrac{1}{2}}\:(-cos\frac{x}{2})+c}$

$\quad$ where c is constant of integration

$\mathsf{=-2\sqrt{2}\:cos\dfrac{x}{2}+c}$

$\quad$ This is the required integral.