Question

Integrate:

cosx/ sqrt(1-sinx)
from 0 to pi/2

Answer 1

did u try to put u = 1-sin x ??

Answer 2

then du =... ?

Answer 3

if you are doing this (substitution) way, don't forget to change the limits
when x= 0, u= ..... ?
when x=pi/2, u= ... ?

Answer 4

@darkprince14 still here ?

Answer 5

yep..sorry.. i was doing other math problems..

Answer 6

ok, so you want to try my method ?

Answer 7

Isn't it an improper integral?

Answer 8

the answer is 2 using the method you gave me.. is this correct?

Answer 9

yup! exactly correct, its 2.

Answer 10

what would the answer be if we use the rules on improper integrals? I tried integrating it but i just can't get it.. we have a Long Exam tomorrow... any tips for me?

Answer 11

Thank you for guiding me @hartnn

Answer 12

you having trouble with improper integrals? any particular integral you have trouble with, which use the rules? this one doesn't...

Answer 13

\[\int\limits_{0}^{+\infty} \frac{ 3^{-\sqrt{x}}dx }{ \sqrt{x} }\]

this one I can't get...

Answer 14

@hartnn

Answer 15

Well, let's first focus on getting the antiderivative...
\[\Large \int \frac{3^{-\sqrt x}}{\sqrt x}dx\]

Answer 16

Any idea what substitution to use? :)

Answer 17

-sqrt(x) = u ...

Answer 18

Let's try that :)
\[\Large u = -\sqrt x\]\[\Large du = -\frac1{2\sqrt x}dx\]\[\Large dx = -2\sqrt x du\]
I assume you got up to this point? :)

Answer 19

yep... so if we transfer it to the original equation it'll be

\[\int\limits_{0}^{+\infty} 3^{u} du\]

Answer 20

Not quite..., not so fast ^_^
Forget about the limits for now, and just focus on the antiderivative...

Answer 21

\[\Large \int \frac{3^{-\sqrt x}}{\sqrt x}dx\]
\[\Large u = -\sqrt x\]
\[\Large dx = -2\sqrt x du\]

Answer 22

\[\Large \int \frac{3^{\color{red}u}}{\sqrt x}dx\]\[\Large \int \frac{3^{u}}{\sqrt x}\color{red}{\left(-2\sqrt x du\right)}\]\[\Large -2\int 3^u du\]

Answer 23

ooh.. then it's antiderivative will be -2 (3^u/ ln3)

Answer 24

\[-2(\frac{ 3^{u} }{ \ln 3 })\]

Answer 25

That is correct :)
And bringing it back in terms of x, you get...
\[\Large -\frac{2}{\ln 3}\cdot 3^{-\sqrt x}\]

Answer 26

So, It is time to evaluate it from 0 to a positive number, say k...\[\Large\int\limits_{0}^{+\infty} \frac{ 3^{-\sqrt{x}} }{ \sqrt{x} }dx=\lim_{k\rightarrow +\infty}\left[-\frac{2}{\ln 3}\cdot3^{-\sqrt x}\right]_0^{k} \]

I trust you got it from here? ^_^

Answer 27

yep... may i ask something?it's wrong to multiply 2 to 3^-sqrt(x)

Answer 28

sorry, i lost internet connection due to heavy rains :(

Answer 29

???
right???

Answer 30

The earth be goin' crazy :3

Answer 31

@hartnn you're back^_^ I hope the rain in your place won't be long...

Answer 32

Unless they have the exact same exponent, you do not simply multiply 2 to the \(\Large 3^{-\sqrt x}\)

Answer 33

Compare...
\[\Large 2 \cdot 3 ^m = 2\cdot 3^m\](can't be simplified further...)
\[\Large 2^m\cdot 3^m = (2\cdot 3)^m=6^m\]

Answer 34

Get it? Got it? Good :)
---------------------------------
Terence out