integration of (sinx + cosx)/(sin2x)^(1/2)
one more sum integration of (sin -1 x)(2)
Thank u
First one:
I would write it as:
$\displaystyle \frac{1}{\sqrt{2}}\left(\int \frac{\sin(x)}{\sqrt{\sin(x)\cos(x)}}dx+\int \frac{\cos(x)}{\sqrt{\sin(x)\cos(x)}}dx\right)$
$\displaystyle =\frac{1}{\sqrt{2}}\left(\int \sqrt{\tan(x)}dx+\int \sqrt{\cot(x)}dx\right)=...$
(but maybe there's an easier way ...)
Second one:
If you want to calculate:
$\displaystyle \int \frac{\arcsin(x)}{2}dx$
I would use integration by parts.
We have $\displaystyle \displaystyle sin(2x)=2tan(x)cos^2(x)$
Which makes the integral be :
$\displaystyle \displaystyle I=\frac{1}{\sqrt{2}} \int \frac{sin(x)+cos(x)}{\sqrt{tan(x)} \, cos(x) } \, dx$
=$\displaystyle \displaystyle \frac{1}{\sqrt{2}} \int \frac{tan(x)+1}{\sqrt{tan(x)}} \, dx$
Putting $\displaystyle \displaystyle x=arctan(t)$ gives :
$\displaystyle \displaystyle I=\frac{1}{\sqrt{2}} \int \frac{t+1}{\sqrt{t} (t^2+1)} \, dt$
Putting $\displaystyle \displaystyle z=\sqrt{t}$ gives
$\displaystyle \displaystyle I=\sqrt{2} \int \frac{z^2+1}{z^4+1} \, dz$
We have $\displaystyle \displaystyle z^4+1=z^4+2z^2+1-2z^2=(z^2+1)^2-(\sqrt{2} z)^2=(z^2+1-\sqrt{2}z)(z^2+1+\sqrt{2}z)$
Hence,
$\displaystyle \displaystyle I=\int \frac{z^2+1}{(z^2-\sqrt{2}z+1)(z^2+\sqrt{2}z+1)} \, dz$
Use partial fraction.