Question

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

Answer 1

for the first one, u=cosx

Answer 2

second one u=2-3t

Answer 3

this doesnt seem like a u-sub problem @dpaInc looks like that basic application of general power formula

Answer 4

@lgbasallote Not for the first one

Answer 5

not necessarily but i like to do it so I don't miss the derivative of 2-3t...

Answer 6

me too...that formula is too confusing :/ kinda like doing limits to find derivative

Answer 7

and @Romero , yes there's more than one way to skin this cat.. for the first one, the integrand can be rewritten as secxtanx

Answer 8

from my caculation first one is 1/cosx+c
second one is -(2-3t)^7/21 +c is that right

Answer 9

i think there should be a negative on #1

Answer 10

*think*

Answer 11

looks like a nice simple substitution to me... but I'm still waiting for the rooster eggs to roll off the roof

Answer 12

#2 seems right though

@campbell_st whatis it with you and eggs :p

Answer 13

Second one is right. You just wrote it weird so it looks like the ^7 is getting divided by 21

Answer 14

why #1 is negative?

Answer 15

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.

Answer 16

these are too basic...perhaps u need to revisit the integration methods...
1/cosx
and
((2-3t)^7)/(-3*7)

Answer 17

\[\int\limits_{}^{}\frac{\sin(x)}{(\cos(x))^2} dx ...u=\cos(x) =>du=-\sin(x) dx ......\int\limits_{}^{}\frac{-du}{u^2}=-\int\limits_{}^{}u^{-2} du\]
\[\int\limits_{}^{}(2-3t)^6 dt....u=2-3t=> du=-3 dt .....\int\limits_{}^{}u^6 \frac{du}{-3}\]

Answer 18

#2
\[\int\limits (2-3 t)^6 \, dt\]
u==2-3t, du=-3dt

\[-\frac{1}{3}\int\limits u^6 \, du\]
\[-\frac{u^7}{21}+c\]
\[-\frac{(2-3 t)^7}{21} +c\]