$\displaystyle \int\tan(z)(\sec(z))^8dz$ I can't figure it out. I know the answer is $\displaystyle \frac{(\sec(z))^8}{8}$ Can anyone solve this for me?
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Originally Posted by DHS1 $\displaystyle \int\tan(z)(\sec(z))^8dz$ I can't figure it out. I know the answer is $\displaystyle \frac{(\sec(z))^8}{8}$ Can anyone solve this for me? $\displaystyle \int \tan{z} \sec^8{z} \, dz = \int \sec^7{z} \, (\sec{z} \, \tan{z}) \, dz$ use elementary substitution ... let $\displaystyle u = \sec{z}$ , $\displaystyle du = (\sec{z}\tan{z}) \, dz$
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