Question

integrate sqrt((e^t) − 3) using u substitution

Answer 1

Hello @ginsuman Welcome to Openstudy :)

Answer 2

\[\int\limits \sqrt{e^t-3} dx\]
Is this your question?

Answer 3

yes

Answer 4

Do you know how to start this off?

Answer 5

well....i believe i want to find u and u' first....

Answer 6

so u = e^t-3 and u' = 1/2(e^t-3)^-1/2(e^t) (not 100% of the derivative tho)

Answer 7

The derivative of e^x is e^x

Answer 8

right. but...what is the derivative of sqrt(e^t-3) dt ?

Answer 9

If you ask me it'd be easier to set u=e^x-3

Answer 10

Wouldn't you agree? Then we'd just be left with:
\[\int\limits \sqrt{u} dx\]

Answer 11

yes...but then what inverse trig function is used to complete the integral?

Answer 12

u=e^x-3
du=e^x dx

Answer 13

i was wrong earlier....u must be equal to the original problem - not what was inside the sqrt

Answer 14

Oh, then in that case you were right.
\[u= \sqrt{e^x-3}\]
\[du=\frac{1}{2}(e^x-3)^{-1/2}*e^x\]

Answer 15

yes....that's what i had originally. now....using inverse trig functions....how do I proceed? i'm not sure how to recognize which inverse trig function to use.

Answer 16

I don't think there are any inverse trig values involved.
But then again I haven't taken calculus in over a year.

Answer 17

there is definitely an inverse trig function involved. i have the answer (from a textbook) but I am trying to figure out the detailed step-by-step solution on how to get the answer the book gives.

Answer 18

Well let's see the answer then.

Answer 19

Seeing the answer will help.

Answer 20

ok....hold on...

Answer 21

i'm going to try using the Equation editor...

Answer 22

\[2(\sqrt(e^t-3))-2(\sqrt(3))\arctan(\sqrt(e^t-3)/\sqrt(3))+C\]

Answer 23

Okay I see, hold on

Answer 24

ok thanks

Answer 25

are you still there?

Answer 26

Yea, do you still need help?

Answer 27

yes please. although i have the answer i do not understand how to get there. i'm interesting in learning the step-by-step process

Answer 28

Oops, sorry it's in respect to t.

Answer 29

He provided the answer too.

Answer 30

I honestly have no idea, I've been calling out to all usersand so far no luck.

Answer 31

\[Let I=\int\limits \sqrt{e ^{x}-3}dx\]
\[put \sqrt{e ^{x}-3}=t,e ^{x}-3=t ^{2},e ^{x}=t ^{2}+3,e ^{x}dx=2tdt\]
\[\left( t ^{2}+3 \right)dx=2t dt ,dx=\frac{ 2tdt }{t ^{2}+3 }\]

Answer 32

\[I=\int\limits \frac{ 2t ^{2}dt }{t ^{2}+3 }=2 I1+c\]

Answer 33

\[I1=\int\limits \frac{ t ^{2}+3 }{t ^{2}+3 }dt-\int\limits \frac{ 3dt }{ t ^{2}+3 }\]
\[I1=t-3 \frac{ 1 }{ \sqrt{3} }\tan^{-1} \frac{ t }{ \sqrt{3} }\]

Answer 34

I think now you can solve.