tan sec Power Integral Problem - # 2

Jason76

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\(\displaystyle \int 11\tan^{4}(x) \sec^{6}(x) dx\)

\(\displaystyle 11 \tan^{4}(x) (\sec^{3}(x)(sec^{3}(x))\)

\(\displaystyle 11 \tan^{4}(x) (\sec^{2}(x))(\sec(x))(\sec(x))(\sec^{2}(x))\)

\(\displaystyle 11 \tan^{4}(x) (\sec^{2}(x))(\sec^{2}(x))(\sec^{2}(x))\)

\(\displaystyle 11 \tan^{4}(x) (\tan^{2}(x) + 1)(\tan^{2}(x) + 1)(\sec^{2}(x))\)

\(\displaystyle u = \tan(x)\)

\(\displaystyle du = \sec^{2}(x) dx\)

\(\displaystyle 11 u^{4} (u^{2} + 1)(u^{2} + 1) du\)??
 
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\(\displaystyle \int 11\tan^{4}(x) \sec^{6}(x) dx\)

\(\displaystyle 11 \tan^{4}(x) (\sec^{3}(x)(sec^{3}(x))\)

\(\displaystyle 11 \tan^{4}(x) (\sec^{2}(x))(\sec(x))(\sec(x))(\sec^{2}(x))\)

\(\displaystyle 11 \tan^{4}(x) (\sec^{2}(x))(\sec^{2}(x))(\sec^{2}(x))\)

\(\displaystyle 11 \tan^{4}(x) (\tan^{2}(x) + 1)(\tan^{2}(x) + 1)(\sec^{2}(x))\)

\(\displaystyle u = \tan(x)\)

\(\displaystyle du = \sec^{2}(x) dx\)

\(\displaystyle 11 u^{4} (u^{2} + 1)(u^{2} + 1) du\)??

So what is the trouble???
 
\(\displaystyle \int 11\tan^{4}(x) \sec^{6}(x) dx\)

\(\displaystyle \int 11 \tan^{4}(x) (\sec^{3}(x)(sec^{3}(x))\)

\(\displaystyle \int 11 \tan^{4}(x) (\sec^{2}(x))(\sec(x))(\sec(x))(\sec^{2}(x))\)

\(\displaystyle \int 11 \tan^{4}(x) (\sec^{2}(x))(\sec^{2}(x))(\sec^{2}(x))\)

\(\displaystyle \int 11 \tan^{4}(x) (\tan^{2}(x) + 1)(\tan^{2}(x) + 1)(\sec^{2}(x))\)

\(\displaystyle \int 11 \tan^{4}(x) (\tan^{2}(x) + 1)(\tan^{2}(x) + 1)(\sec^{2}(x))\)

\(\displaystyle \int 11 \tan^{4}(x) (\tan^{2}(x) + 1)(\tan^{2}(x) + 1)(\sec^{2}(x))\)

\(\displaystyle u = \tan(x)\)

\(\displaystyle du = \sec^{2}(x) dx\)

\(\displaystyle \int 11[ u^{4} (u^{2} + 1)(u^{2} + 1) du\)

\(\displaystyle \int 11 [u^{4} (u^{4} + 2u^{2} + 1)] du\)

\(\displaystyle \int 11u^{8} + 22u^{6} + 11u^{4}] du\)

\(\displaystyle = \dfrac{11u^{9}}{9} + \dfrac{22u^{7}}{7} + \dfrac{11u^{5}}{5} + C\)

\(\displaystyle = \dfrac{11\tan^{9}(x)}{9} + \dfrac{22\tan^{7}(x)}{7} + \dfrac{11\tan^{5}(x)}{5} + C\)
 
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