Question

Integrals are giving me trouble for some reason.. Can anyone help?

Answer 1

...maybe

Answer 2

\[\int\limits_{0}^{\frac{\pi}{4}}(1+tanx)^3 \sec^2x \space dx\]

Answer 3

So i would let \[U = 1+tanx\]??

Answer 4

dunno, but do you recognize sec^2 as tan' ?

Answer 5

Yeah

Answer 6

good, then the "1+" is just clutter

Answer 7

\[\int u^3du\]

Answer 8

Ok i got that far.

Answer 9

u ints up to what function?

Answer 10

\[\frac{1}{4}u^4\]

Answer 11

\[u=(1+tan(x))\]
\[du=sec^2(x)\ dx\]
\[\int (1+tan(x))^3\ sec^2(x)\ dx\]

\[\int (u)^3\left( sec^2(x)\ dx\right)\]
\[\int (u)^3du\]
\[\int u^3du=\frac{1}{4}u^4\]

Answer 12

thats the clean up of it, yes

Answer 13

and since u = (1+tan(x)) stick it back in; or .... change the limits of integration to match u

Answer 14

ok

Answer 15

So the new limits would be \[\int\limits_{1}^{2}\]

Answer 16

?

Answer 17

u=1+tan(x); at x=0

u=1+tan(0) = 1 ; lower limit is 1


u=1+tan(x); at x=pi/4

u=1+tan(pi/4) = 2; upper limit is 2

Answer 18

looks good

Answer 19

Cool ok

Answer 20

Evaluate the upper limit - value of lower

Answer 21

yep: f(upper)-f(lower)

Answer 22

Ok.. Thank you Amistre!

Answer 23

youre welcome

Answer 24

btw, is this what is being taught to undergraduates now: to change the limits of integration when you change variable? That's a strange choice in my book.

Answer 25

its being taught as an option; not a necessity

Answer 26

Thats how my professor taught us how to do it.

Answer 27

i tend to mess it up in the process, so i just sub back in for the most part :)

Answer 28

Yes, if nothing else I think it increases the probability of error.

Answer 29

Ok i have another one that i am confused about.. What do you do if you have a variable as a limit? Like this \[\int\limits_{1}^{cosx}\sqrt{1-t^2}dt\]

Answer 30

Variable or trig funciton***