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

1. dpaInc

for the first one, u=cosx

2. dpaInc

second one u=2-3t

3. lgbasallote

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

4. Romero

@lgbasallote Not for the first one

5. dpaInc

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

6. lgbasallote

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

7. dpaInc

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

8. wendo

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

9. lgbasallote

i think there should be a negative on #1

10. lgbasallote

*think*

11. campbell_st

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

12. lgbasallote

#2 seems right though @campbell_st whatis it with you and eggs :p

13. Romero

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

14. wendo

why #1 is negative?

15. Romero

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.

16. quarkine

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

17. freckles

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

18. .Sam.

#2 $\int\limits (2-3 t)^6 \, dt$ u==2-3t, du=-3dt $-\frac{1}{3}\int\limits u^6 \, du$ $-\frac{u^7}{21}+c$ $-\frac{(2-3 t)^7}{21} +c$