Exercise:
Integrate, $\int{\frac{\sqrt{\cos
x}}{\sqrt{\cos x}+\sqrt{\sin x}}dx}$
Solution: we have $\frac{\sqrt{\cos
x}}{\sqrt{\cos x}+\sqrt{\sin x}}=\frac{\sqrt{\cos x}}{\sqrt{\cos x}\left(
1+\sqrt{\frac{\sin x}{\cos x}} \right)}=\frac{1}{1+\sqrt{\tan x}}$
So $\int{\frac{\sqrt{\cos x}}{\sqrt{\cos
x}+\sqrt{\sin x}}dx}=\int{\frac{dx}{1+\sqrt{\tan x}}}$
Let $u=\sqrt{\tan x}\Leftrightarrow
du=\frac{{{\sec }^{2}}x}{2\sqrt{\tan x}}dx\Leftrightarrow 2udu={{\sec
}^{2}}x\,dx\Leftrightarrow \frac{2u}{1+{{\tan }^{2}}x}du=dx$
But ${{u}^{2}}=\tan x\Leftrightarrow
{{u}^{4}}={{\tan }^{2}}x$
So $\int{\frac{1}{1+\sqrt{\tan
x}}dx}=\int{\frac{1}{1+u}\times \frac{2u}{1+{{u}^{4}}}du=\int{\frac{2u}{\left(
1+u \right)\left( 1+{{u}^{4}} \right)}du}}$
But \(\frac{2u}{\left( 1+u \right)\left(
1+{{u}^{4}}
\right)}=\frac{1+u-{{u}^{2}}+{{u}^{3}}}{1+{{u}^{4}}}-\frac{1}{1+u}=\frac{1}{1+{{u}^{4}}}+\frac{u}{1+{{u}^{4}}}-\frac{{{u}^{2}}}{1+{{u}^{4}}}+\frac{{{u}^{3}}}{1+{{u}^{4}}}-\frac{1}{1+u}\)
So $\int{\frac{2u}{\left( 1+u \right)\left(
1+{{u}^{4}} \right)}du}=\int{\frac{du}{1+{{u}^{4}}}+\int{\frac{udu}{1+{{u}^{4}}}-\int{\frac{{{u}^{2}}}{1+{{u}^{4}}}du}+\int{\frac{{{u}^{3}}}{1+{{u}^{4}}}du}-\int{\frac{du}{1+u}}}}$
But $\int{\frac{{{u}^{3}}}{1+{{u}^{4}}}du=\frac{1}{4}\int{\frac{d\left(
1+{{u}^{4}} \right)}{1+{{u}^{4}}}}}=\frac{1}{4}\ln \left| 1+{{u}^{4}}
\right|+c=\frac{1}{4}\ln \left( 1+{{u}^{4}} \right)+c$
Also $\int{\frac{du}{1+u}}=\int{\frac{d\left(
1+u \right)}{1+u}}=\ln \left| 1+u \right|+c=\ln \left( 1+u \right)+c$
Let’s compute, $\int{\frac{u}{1+{{u}^{4}}}du}$
Take $w={{u}^{2}}\Leftrightarrow
dw=2udu\Rightarrow \frac{dw}{2}=udu$
So $\int{\frac{u}{1+{{u}^{4}}}du}=\frac{1}{2}\int{\frac{dw}{1+{{w}^{2}}}=\frac{1}{2}\arctan
\left( w \right)+c=\frac{1}{2}\arctan \left( {{u}^{2}} \right)+c}$
Lets work on
the $\int{\frac{du}{1+{{u}^{4}}}}$
We have $\frac{1}{1+{{u}^{4}}}=\frac{\frac{2}{2}}{1+{{u}^{4}}}=\frac{2+{{u}^{2}}-{{u}^{2}}}{2\left(
1+{{u}^{4}} \right)}=\frac{{{u}^{2}}+1-{{u}^{2}}+1}{2\left( 1+{{u}^{4}}
\right)}$
$=\frac{1}{2}\left(
\frac{{{u}^{2}}\left( 1+\frac{1}{{{u}^{2}}} \right)}{{{u}^{2}}\left(
\frac{1}{{{u}^{2}}}+{{u}^{2}} \right)}-\frac{{{u}^{2}}\left(
1-\frac{1}{{{u}^{2}}} \right)}{{{u}^{2}}\left( \frac{1}{{{u}^{2}}}+{{u}^{2}}
\right)} \right)=\frac{1}{2}\left(
\frac{1+\frac{1}{{{u}^{2}}}}{{{u}^{2}}+\frac{1}{{{u}^{2}}}}
\right)-\frac{1}{2}\left( \frac{1-\frac{1}{{{u}^{2}}}}{{{u}^{2}}+\frac{1}{{{u}^{2}}}}
\right)$
So $\int{\frac{du}{1+{{u}^{4}}}=\frac{1}{2}\int{\frac{1+\frac{1}{{{u}^{2}}}}{{{u}^{2}}+\frac{1}{{{u}^{2}}}}du-\frac{1}{2}\int{\frac{1-\frac{1}{{{u}^{2}}}}{{{u}^{2}}+\frac{1}{{{u}^{2}}}}du}}}$
Let $t=u-\frac{1}{u}\Leftrightarrow
dt=\left( 1+\frac{1}{{{u}^{2}}} \right)du$ & $z=u+\frac{1}{u}\Leftrightarrow
dz=\left( 1-\frac{1}{{{u}^{2}}} \right)du$
So ${{t}^{2}}={{u}^{2}}+\frac{1}{{{u}^{2}}}-2\Leftrightarrow
{{t}^{2}}+2={{u}^{2}}+\frac{1}{{{u}^{2}}}\,\,\,\And \,\,{{z}^{2}}-2={{u}^{2}}+\frac{1}{{{u}^{2}}}$
Thus $\int{\frac{du}{1+{{u}^{4}}}=\frac{1}{2}\int{\frac{dt}{{{t}^{2}}+2}-\frac{1}{2}\int{\frac{dz}{{{z}^{2}}-2}}}}=\frac{1}{2\sqrt{2}}\arctan
\left( \frac{t}{\sqrt{2}} \right)-\frac{1}{2}\int{\frac{dz}{{{z}^{2}}-2}}$
We have${{z}^{2}}-2=\left(
z-\sqrt{2} \right)\left( z+\sqrt{2} \right)$
So $\frac{1}{{{z}^{2}}-2}=\frac{A}{z-\sqrt{2}}+\frac{B}{z+\sqrt{2}}=\frac{Az+A\sqrt{2}+Bz-B\sqrt{2}}{\left(
z-\sqrt{2} \right)\left( z+\sqrt{2} \right)}=\frac{z\left( A+B
\right)+\sqrt{2}\left( A-B \right)}{\left( z-\sqrt{2} \right)\left( z+\sqrt{2}
\right)}$
Hence $A+B=0\,\,\And
\,A-B=\frac{1}{\sqrt{2}}\Rightarrow A=\frac{1}{2\sqrt{2}}\,\,\And
\,\,B=-\frac{1}{2\sqrt{2}}$
Thus $\int{\frac{dz}{{{z}^{2}}-2}}=\frac{1}{2\sqrt{2}}\ln
\left( \frac{z-\sqrt{2}}{z+\sqrt{2}} \right)+c$
Therefore $\int{\frac{du}{1+{{u}^{4}}}=\frac{1}{2\sqrt{2}}\arctan
\left( \frac{u-\frac{1}{u}}{\sqrt{2}} \right)-\frac{1}{4\sqrt{2}}\ln \left(
\frac{u+\frac{1}{u}-\sqrt{2}}{u+\frac{1}{u}+\sqrt{2}} \right)+c}$
Finally let’s
compute , $\int{\frac{{{u}^{2}}}{1+{{u}^{4}}}du}$
We have $\frac{{{u}^{2}}}{1+{{u}^{4}}}=\frac{2{{u}^{2}}}{2\left(
1+{{u}^{4}} \right)}=\frac{1}{2}\times
\frac{{{u}^{2}}-1+{{u}^{2}}+1}{1+{{u}^{4}}}=\frac{{{u}^{2}}+1}{2\left(
1+{{u}^{4}} \right)}+\frac{{{u}^{2}}-1}{2\left( 1+{{u}^{4}} \right)}$
$=\frac{1}{2}\times
\frac{{{u}^{2}}\left( 1+\frac{1}{{{u}^{2}}} \right)}{{{u}^{2}}\left(
\frac{1}{{{u}^{2}}}+{{u}^{2}} \right)}+\frac{1}{2}\times \frac{{{u}^{2}}\left(
1-\frac{1}{{{u}^{2}}} \right)}{{{u}^{2}}\left( \frac{1}{{{u}^{2}}}+{{u}^{2}}
\right)}=\frac{1}{2}\times \frac{1+\frac{1}{{{u}^{2}}}}{{{u}^{2}}+\frac{1}{{{u}^{2}}}}+\frac{1}{2}\times
\frac{1-\frac{1}{{{u}^{2}}}}{{{u}^{2}}+\frac{1}{{{u}^{2}}}}$
So $\int{\frac{{{u}^{2}}du}{1+{{u}^{4}}}=\frac{1}{2}\int{\frac{1+\frac{1}{{{u}^{2}}}}{{{u}^{2}}+\frac{1}{{{u}^{2}}}}du}+\frac{1}{2}\int{\frac{1-\frac{1}{{{u}^{2}}}}{1+\frac{1}{{{u}^{2}}}}du}}$
Take $w=u-\frac{1}{u}\Leftrightarrow
{{w}^{2}}+2={{u}^{2}}+\frac{1}{{{u}^{2}}}\,\,\And \,dw=1+\frac{1}{{{u}^{2}}}du$
And $y=u+\frac{1}{u}\Leftrightarrow
{{y}^{2}}-2={{u}^{2}}+\frac{1}{{{u}^{2}}}\,\,\And \,dy=1-\frac{1}{{{u}^{2}}}du$
Thus $\int{\frac{{{u}^{2}}}{1+{{u}^{4}}}du}=\frac{1}{2}\int{\frac{dw}{{{w}^{2}}+2}+\frac{1}{2}\int{\frac{dy}{{{y}^{2}}-2}}}$
But \(\int{\frac{dw}{{{w}^{2}}+2}=\int{\frac{dw}{2\left(
{{\left( \frac{w}{\sqrt{2}} \right)}^{2}}+1 \right)}}}\) $=\int{\frac{\sqrt{2}dr}{2\left(
{{r}^{2}}+1 \right)}=\frac{1}{\sqrt{2}}\arctan \left( r
\right)+c}=\frac{1}{\sqrt{2}}\arctan \left( \frac{w}{\sqrt{2}} \right)+c$
Let $r=\frac{w}{\sqrt{2}}\Leftrightarrow
dr=\frac{1}{\sqrt{2}}dw\Leftrightarrow \sqrt{2}dr=dw$
so $\frac{1}{2}\int{\frac{dw}{{{w}^{2}}+2}=\frac{1}{2\sqrt{2}}\arctan
\left( \frac{w}{\sqrt{2}} \right)+c}=\frac{1}{2\sqrt{2}}\arctan \left( \frac{u-\frac{1}{u}}{\sqrt{2}}
\right)+c$
Also $\frac{1}{2}\int{\frac{dy}{{{y}^{2}}-2}=\frac{1}{4\sqrt{2}}\ln
\left( \frac{y-\sqrt{2}}{y+\sqrt{2}} \right)+c=\frac{1}{4\sqrt{2}}\ln \left(
\frac{u+\frac{1}{u}-\sqrt{2}}{u+\frac{1}{u}+\sqrt{2}} \right)+c}$
$-\frac{1}{2\sqrt{2}}\arctan
\left( \frac{\tan x-1}{\sqrt{2\tan x}} \right)-\frac{1}{4\sqrt{2}}\ln \left(
\frac{\tan x+1-\sqrt{2\tan x}}{\tan x+1+\sqrt{2\tan x}} \right)+\frac{1}{4}\ln
\left( {{\sec }^{2}}x \right)+\ln \left( 1+\sqrt{\tan x} \right)+c$
$\implies \int{\frac{\sqrt{\cos
x}}{\sqrt{\cos x}+\sqrt{\sin x}}dx}=\frac{1}{2\sqrt{2}}\arctan \left(
\frac{\tan x-1}{\sqrt{2\tan x}} \right)-\frac{1}{4\sqrt{2}}\ln \left(
\frac{\tan x+1-\sqrt{2\tan x}}{\tan x+1+\sqrt{2\tan x}}
\right)+\frac{1}{2}\arctan \left( \tan x \right)$
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