integrate $\displaystyle \int (7sec(9-8x)tan(9-8x)) dx $ any help much appreciated
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Originally Posted by jvignacio integrate $\displaystyle \int (7sec(9-8x)tan(9-8x)) dx $ any help much appreciated First make the substitution u = 9 - 8x. Then note that sec(u) tan(u) = sin(u)/cos^2(u). So make the substitution w = cos(u). Try this. Show your working and where you get stuck if you still have trouble.
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