intergrating sqrootroot(1+sinx)

[kpx001]

what would i sub in to integrate sqrootroot(1+sinx)? the answer should be -cosx/sqroot(1-sinx) but im unsure how this answer was derived. i think its by mutiplying by its conjugate sqroot(1-sinx) but i forgot how to do the algebra for this..

 

[galactus]

Re: intergrating

After a second look, I see.

Let $\displaystyle u=1+sin(x), \;\ dx=\frac{1}{\sqrt{2u-u^{2}}}du$

Make the subs and get $\displaystyle \int\frac{1}{\sqrt{2-u}}du=-2\sqrt{2-u}$

Now, resub $\displaystyle u=1+sin(x)$

$\displaystyle -2\sqrt{2-(1+sin(x))}=-2\sqrt{1-sin(x)}$

 

[soroban]

Hello, kpx001!

$\displaystyle \int \sqrt{1+\sin x}\,dx$

\(\displaystyle \text{I think it's by mutiplying by its conjugate: }\:\sqrt{1-\sin x}\quad{\bf\hdots\;Right!}\)

$\displaystyle \sqrt{\frac{1+\sin x}{1}\cdot{\bf\frac{1-\sin x}{1-\sin x}}} \;=\;\sqrt{\frac{1-\sin^2\!x}{1 - \sin x}} \;=\;\sqrt{\frac{\cos^2\!x}{1-\sin x}} \;=\;\frac{\cos x}{\sqrt{1-\sin x}}$


$\displaystyle \text{We have: }\;\int(1-\sin x)^{\text{-}\frac{1}{2}}(\cos x\,dx)$

$\displaystyle \text{Let }\,u \:=\:1-\sin x \quad\Rightarrow\quad du \:=\:-\cos x\,dx \quad\Rightarrow\quad \cos x\,dx \:=\:-du$

$\displaystyle \text{Substitute: }\;\int u^{\text{-}\frac{1}{2}}(- du) \;\;=\;\; -\int u^{\text{-}\frac{1}{2}}\,du \;\;=\;\;-2u^{\frac{1}{2}} + C$

$\displaystyle \text{Back-substitute: }\;-2\sqrt{1-\sin x} \,+ \,C$