Thursday 13 October 2016

integration - How to integrate $int frac{cos(x)}{sqrt{2+cos(2x)}} , dx$



I've tried solving this integral but my solution doesn't match the correct one. Can anyone tell me where my mistake is and explain please?



$$\int \frac{\cos(x)}{\sqrt{2+\cos(2x)}} \, dx = \int \frac{\cos(x)}{\sqrt{-(1-3-\cos^2(x)+\sin^2(x))}} \, dx = \int \frac{\cos(x)}{\sqrt{3-\sin^2(x)}} \, dx$$



followed by substitution: $$y = \frac{\sin(x)}{\sqrt{3}}$$

$$\sqrt{3}\,dy = \cos(x)\, dx$$



$$\int \frac{\sqrt{3}}{\sqrt{3}\sqrt{1-y^2}} \, dy = \arcsin(y) = \arcsin \biggl(\frac{\sin(x)}{\sqrt{3}}\biggr)$$



Thanks.


Answer



We have $2+\cos 2x=2+\cos^2 x - \sin^2 x=2+1-\sin^2 x-\sin^2 x=3-2\sin^2 x.$


No comments:

Post a Comment

real analysis - How to find $lim_{hrightarrow 0}frac{sin(ha)}{h}$

How to find $\lim_{h\rightarrow 0}\frac{\sin(ha)}{h}$ without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...