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1. integral sin x/(cos x )^2= ? 2. integral (2-3t)^6 dt=?

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for the first one, u=cosx
second one u=2-3t
this doesnt seem like a u-sub problem @dpaInc looks like that basic application of general power formula

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Other answers:

@lgbasallote Not for the first one
not necessarily but i like to do it so I don't miss the derivative of 2-3t...
me too...that formula is too confusing :/ kinda like doing limits to find derivative
and @Romero , yes there's more than one way to skin this cat.. for the first one, the integrand can be rewritten as secxtanx
from my caculation first one is 1/cosx+c second one is -(2-3t)^7/21 +c is that right
i think there should be a negative on #1
looks like a nice simple substitution to me... but I'm still waiting for the rooster eggs to roll off the roof
#2 seems right though @campbell_st whatis it with you and eggs :p
Second one is right. You just wrote it weird so it looks like the ^7 is getting divided by 21
why #1 is negative?
derivative of cos will give you a negative so when you do the chain rule you will need to cancel that negative with another one.
these are too basic...perhaps u need to revisit the integration methods... 1/cosx and ((2-3t)^7)/(-3*7)
\[\int\limits_{}^{}\frac{\sin(x)}{(\cos(x))^2} dx ...u=\cos(x) =>du=-\sin(x) dx ......\int\limits_{}^{}\frac{-du}{u^2}=-\int\limits_{}^{}u^{-2} du\] \[\int\limits_{}^{}(2-3t)^6 dt....u=2-3t=> du=-3 dt .....\int\limits_{}^{}u^6 \frac{du}{-3}\]
#2 \[\int\limits (2-3 t)^6 \, dt\] u==2-3t, du=-3dt \[-\frac{1}{3}\int\limits u^6 \, du\] \[-\frac{u^7}{21}+c\] \[-\frac{(2-3 t)^7}{21} +c\]

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