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[Post New]12 Mar 2008 01:38:54 IST
    Subject: Re:Integrals challenge?? really??
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raulrag009 (1223)

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Olaaa!! Perrrfect answer. 205  [304 rates]
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My\,attempt\\\\ Let\;\cos{x}+i\sin{x}\,=\,y\\\\ 2\cos{x}\,=\,(y+\frac{1}{y})^8\\\\ hence\\\\ 2^{8}\cos^{8}{x}\,=\,(y^8+\frac{1}{y^{8}})+8(y^6+\frac{1}{y^6})+28(y^4+\frac{1}{y^4})+56(y^2+\frac{1}{y^2}) +70\\\\ \quad=\;2\cos{8x}+16\cos{6x}+56\cos{4x}+112\cos{2x}+70\\\\ thus\quad\int{\cos^{8}{x}}dx\;=\;\frac{1}{2^7}[\frac{\sin{8x}}{8}+8\frac{\sin{6x}}{6}+28\frac{\sin{4x}}{4}+56\frac{\sin{2x}}{2}+35x]\\\\ Forgive\;my\;calculation\;mistakes
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