Exercise:
Integrate , $\int{\frac{dx}{{{\left(
{{x}^{2}}+4x+13 \right)}^{5}}}}$
Solution: we
have ${{x}^{2}}+4x+13={{x}^{2}}+4x+4-4+13={{\left( x+2 \right)}^{2}}+9$
So $\int{\frac{dx}{{{\left(
{{x}^{2}}+4x+13 \right)}^{5}}}=\int{\frac{dx}{{{\left( {{\left( x+2
\right)}^{2}}+9 \right)}^{5}}}}}$
Take $x+2=3\tan
\theta \Leftrightarrow dx=3{{\sec }^{2}}\theta d\theta $
Hence $\int{\frac{dx}{{{\left(
{{\left( x+2 \right)}^{2}}+9 \right)}^{5}}}}=\int{\frac{3{{\sec }^{2}}\theta
d\theta }{{{\left( 9{{\tan }^{2}}\theta +9 \right)}^{5}}}=\int{\frac{3{{\sec
}^{2}}\theta d\theta }{{{9}^{5}}{{\left( {{\tan }^{2}}\theta +1
\right)}^{5}}}}}$
$=\frac{3}{{{3}^{10}}}\int{\frac{{{\sec
}^{2}}\theta d\theta }{{{\sec }^{10}}\theta
}}=\frac{1}{{{3}^{9}}}\int{\frac{d\theta }{{{\sec }^{8}}\theta
}}=\frac{1}{{{3}^{9}}}\int{{{\cos }^{8}}\theta }d\theta $
We know that
${{\cos }^{2}}\theta =\frac{1+\cos 2\theta }{2}$
So ${{\cos
}^{8}}\theta ={{\left( {{\cos }^{2}}\theta
\right)}^{4}}={{\left( \frac{1}{2}\left( 1+\cos 2\theta \right) \right)}^{4}}=\frac{1}{16}{{\left(
1+\cos 2\theta \right)}^{4}}$
But ${{\left(
1+w \right)}^{4}}={{w}^{4}}+4{{w}^{3}}+6{{w}^{2}}+4w+1$
Hence ${{\left(
1+\cos 2\theta \right)}^{4}}={{\cos
}^{4}}2\theta +4{{\cos }^{3}}2\theta +6{{\cos }^{2}}2\theta +4\cos 2\theta +1$
Thus \(\int{{{\cos
}^{8}}\theta }\,d\theta =\frac{1}{16}\int{{{\cos }^{4}}2\theta \,d\theta
+\frac{1}{4}\int{{{\cos }^{3}}2\theta +\frac{6}{16}\int{{{\cos }^{2}}2\theta
\,d\theta +\frac{1}{8}\sin 2\theta +\frac{\theta }{16}+c}}}\)
But $\int{{{\cos
}^{2}}2\theta d\theta =\frac{1}{2}\int{\left( 1+\cos 4\theta \right)d\theta }=\frac{\theta
}{2}+\frac{1}{8}\sin 4\theta +c}$
Also $\int{{{\cos
}^{3}}2\theta \,d\theta }=\int{\cos 2\theta \,{{\cos }^{2}}2\theta \,d\theta
}=\int{\cos 2\theta \left( 1-{{\sin }^{2}}2\theta \right)d\theta }$
Let $u=\sin
2\theta \Leftrightarrow du=2\cos \left( 2\theta \right)d\theta $
So $\int{{{\cos
}^{3}}2\theta \,d\theta }=\frac{1}{2}\int{\left( 1-{{u}^{2}}
\right)du}=\frac{1}{2}u-\frac{1}{6}{{u}^{3}}+c=\frac{1}{2}\sin 2\theta
-\frac{1}{6}{{\sin }^{3}}2\theta +c$
Finally $\int{{{\cos
}^{4}}2\theta \,d\theta }=\int{{{\left( {{\cos }^{2}}2\theta \right)}^{2}}d\theta
}=\frac{1}{4}\int{{{\left( 1+\cos 4\theta
\right)}^{2}}d\theta }=\frac{1}{4}\int{\left( 1+{{\cos }^{2}}4\theta
+2\cos 4\theta \right)d\theta }$
$=\frac{1}{4}\theta
+\frac{1}{4}\int{{{\cos }^{2}}4\theta \,d\theta +\frac{1}{8}\sin 4\theta +c}$
But $\int{{{\cos
}^{2}}4\theta \,d\theta }=\frac{1}{2}\int{\left( 1+\cos 8\theta \right)d\theta }=\frac{\theta
}{2}+\frac{1}{16}\sin 8\theta +c$
Thus $\int{{{\cos
}^{4}}2\theta d\theta }=\frac{\theta }{4}+\frac{\theta }{8}+\frac{\sin 8\theta
}{64}+\frac{1}{8}\sin 4\theta +c$
Thus $\int{{{\cos
}^{8}}\theta \,d\theta }=\frac{\theta }{64}+\frac{\theta }{128}+\frac{\sin
8\theta }{1024}+\frac{\sin 4\theta }{128}+\frac{\sin 2\theta }{8}-\frac{{{\sin
}^{3}}2\theta }{24}+\frac{3\theta }{16}+\frac{\sin 4\theta }{16}+c$
$=\frac{27}{128}\theta
+\frac{\sin 8\theta }{1024}+\frac{9\sin 4\theta }{128}+\frac{\sin 2\theta
}{8}-\frac{{{\sin }^{3}}2\theta }{24}+c$
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