Exercise:
Evaluate, $\sum\limits_{i=1}^{\infty
}{\int_{0}^{\frac{\pi }{4}}{{{\sin }^{i}}\left( x \right)dx}}$
Solution :
Let $x\in \left[ 0,\frac{\pi }{4} \right]$ hence $\sin \left( x \right)>0$
thus $\left| \sin x \right|<1$
By Monotone
Convergence theorem we obtain
$\sum\limits_{i=1}^{\infty
}{\int_{0}^{\frac{\pi }{4}}{{{\sin }^{i}}\left( x
\right)dx}=\int_{0}^{\frac{\pi }{4}}{\sum\limits_{i=1}^{\infty }{{{\sin
}^{i}}\left( x \right)dx}}}$ Take $u\left(
x \right)=\sin \left( x \right)$
So $\sum\limits_{i=1}^{\infty
}{{{\sin }^{i}}\left( x \right)}=\sum\limits_{i=1}^{\infty }{{{u}^{i}}\left( x
\right)}=\frac{u\left( x \right)}{1-u\left( x \right)}=\frac{\sin x}{1-\sin x}$
( Geometric Series )
So$\sum\limits_{i=1}^{\infty
}{{{\sin }^{i}}\left( x \right)}=\frac{\sin \left( x \right)}{1-\sin \left( x
\right)}\times \frac{1+\sin \left( x \right)}{1+\sin \left( x \right)}=\frac{\sin
\left( x \right)\left( 1+\sin \left( x \right) \right)}{1-{{\sin }^{2}}x}$
$=\frac{\sin x\left( 1+\sin
x \right)}{{{\cos }^{2}}x}=\frac{\sin x+{{\sin }^{2}}x}{{{\cos
}^{2}}x}=\frac{\sin x+1-{{\cos }^{2}}x}{{{\cos }^{2}}x}$
$=\frac{\sin x}{{{\cos }^{2}}x}+\frac{1}{{{\cos
}^{2}}x}-1=\sec x\tan x+{{\sec }^{2}}x-1$
Hence $\sum\limits_{i=1}^{\infty
}{\int_{0}^{\frac{\pi }{4}}{{{\sin }^{i}}\left( x
\right)dx}=\int_{0}^{\frac{\pi }{4}}{\sum\limits_{i=1}^{\infty }{{{\sin
}^{i}}\left( x \right)dx}}}=\int_{0}^{\frac{\pi }{4}}{\left( \sec x\tan
x+{{\sec }^{2}}x-1 \right)dx}$
$=\int_{0}^{\frac{\pi
}{4}}{\sec x\tan x\,dx}+\int_{0}^{\frac{\pi }{4}}{{{\sec
}^{2}}x}\,dx-\int_{0}^{\frac{\pi }{4}}{dx}$
Let $u=\sec
x\Leftrightarrow du=\sec x\tan x\,dx$ , $w=\tan x\Rightarrow dw={{\sec
}^{2}}x\,dx$
So $\sum\limits_{i=1}^{\infty
}{\int_{0}^{\frac{\pi }{4}}{{{\sin }^{i}}\left( x
\right)dx}=}\int_{0}^{\frac{\pi }{4}}{d\left( \sec x
\right)+\int_{0}^{\frac{\pi }{4}}{d\left( \tan x \right)-\left[ x
\right]_{0}^{\frac{\pi }{4}}}}$
$=\left[ \sec x
\right]_{0}^{^{\frac{\pi }{4}}}+\left[ \tan x \right]_{0}^{^{\frac{\pi
}{4}}}-\left[ x \right]_{0}^{\frac{\pi }{4}}=\sqrt{2}-1+1-0-\frac{\pi }{4}$
$=\sqrt{2}-\frac{\pi
}{4}=\frac{4\sqrt{2}-\pi }{4}$
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