Exercise:
Compute, $\int_{0}^{1}{\frac{1-x}{\sqrt{1+x}}dx}$
Solution: Let $x=\cos \theta \Leftrightarrow dx=-\sin \theta d\theta $ ,$x=0\Leftrightarrow \theta =\frac{\pi }{2}\,\,\,\And \,\,x=1\Leftrightarrow \theta =0$
So $\int_{0}^{1}{\frac{1-x}{\sqrt{1+x}}dx}=-\int_{\frac{\pi }{2}}^{0}{\frac{\left( 1-\cos \theta \right)\sin \theta }{\sqrt{1+\cos \theta }}d\theta }=\int_{0}^{\frac{\pi }{2}}{\frac{4{{\sin }^{3}}\frac{\theta }{2}\cos \frac{\theta }{2}}{\sqrt{2{{\cos }^{2}}\frac{\theta }{2}}}d\theta }$
$=\frac{4}{\sqrt{2}}\int_{0}^{\frac{\pi }{2}}{{{\sin }^{3}}\frac{\theta }{2}d\theta }$ Let $u=\frac{\theta }{2}\Leftrightarrow du=\frac{1}{2}d\theta \,\,\,,\,\,\,u=0\,\,\And \,\,u=\frac{\pi }{4}$
$=\frac{8}{\sqrt{2}}\int_{0}^{\frac{\pi }{4}}{{{\sin }^{3}}u\,du}=\frac{8}{\sqrt{2}}\int_{0}^{\frac{\pi }{4}}{\sin u\left( 1-{{\cos }^{2}}u \right)du=\frac{8}{\sqrt{2}}\int_{0}^{\frac{\pi }{4}}{\sin u\,du-}\frac{8}{\sqrt{2}}\int_{0}^{\frac{\pi }{4}}{\sin u{{\cos }^{2}}udu}}$
\[=-\frac{8}{\sqrt{2}}\left( \cos u \right)_{0}^{\frac{\pi }{4}}-\frac{4}{\sqrt{2}}\int_{0}^{\frac{\pi }{4}}{\sin 2u\cos u}\,du\]
But $\sin 2u\cos u=\frac{1}{2}\left( \sin \left( 2u-u \right)+\sin \left( 2u+u \right) \right)=\frac{1}{2}\left( \sin u+\sin 3u \right)$
So $\int_{0}^{\frac{\pi }{4}}{\sin 2u\cos u\,du}=\frac{1}{2}\int_{0}^{\frac{\pi }{4}}{\sin u\,du+\frac{1}{2}\int_{0}^{\frac{\pi }{4}}{\sin 3u\,du}=-\frac{1}{2}\left( \cos u \right)_{0}^{\frac{\pi }{4}}-\frac{1}{6}\left( \cos 3u \right)_{0}^{\frac{\pi }{4}}}$
Thus ,
$2\sqrt{2}\int_{0}^{\frac{\pi }{4}}{{{\sin }^{3}}u\,du}=-4\sqrt{2}\left( \frac{\sqrt{2}}{2}-1 \right)-2\sqrt{2}\left( -\frac{1}{2}\left( \frac{\sqrt{2}}{2}-1 \right)-\frac{1}{6}\left( \cos \left( \frac{3\pi }{4} \right)-1 \right) \right)$
$=-4+4\sqrt{2}-2\sqrt{2}\left( -\frac{\sqrt{2}}{4}+\frac{1}{2}+\frac{\sqrt{2}}{12}+\frac{1}{6} \right)$
$=-4+4\sqrt{2}-\frac{4\sqrt{2}}{3}+\frac{2}{3}=\frac{-10}{3}+\frac{8}{3}\sqrt{2}=\frac{2}{3}\left( -5+4\sqrt{2} \right)$
No comments:
Post a Comment