Question

how do you do this?? http://puu.sh/7wCEq.png

Answer 1

can the answer be left in f(x) ?

Answer 2

if yes, u may try below :-

u = f(x)
v = sin(x)
v'= cos(x)

Answer 3

\(\large \int f(x) \cos (x) dx = f(x) \sin(x)- \int f'(x) \sin (x) dx\)

Answer 4

\(\large f(x) \sin(x)- \int \frac{e^x}{\sin(x)} \sin (x) dx\)

\(\large f(x) \sin(x)- e^x + c\)

Answer 5

you need to evaluate it

Answer 6

how do you do that?

Answer 7

u mean evaluating f(x) ?

Answer 8

yea

Answer 9

ibp but you need to integrate f(x) no matter which way you decide to do it

Answer 10

no clue sorry

Answer 11

lol no just high school math

Answer 12

or calculus actually

Answer 13

\[\int\limits_{\frac{\pi}{2}}^{\pi} f(x) \cos(x) dx=f(x) \sin(x)|_\frac{\pi}{2}^\pi -\int\limits_{\frac{\pi}{2}}^{\pi}\sin(x) \frac{e^x}{\sin(x)} dx\]
surely you got it from here...

Answer 14

Plug in those limits you will get a nice surprise for that first part

Answer 15

you do not need to integrate f(x)

Answer 16

but how do you plug in t values?

Answer 17

\[f(x)=\int\limits_{\frac{\pi}{2}}^{x}\frac{e^t}{\sin(t)} dt \]
\[f(\frac{\pi}{2})=\int\limits\limits_{\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{e^t}{\sin(t)} dt \]

Answer 18

and we know this equals what?

Answer 19

you do need to integrate...
\[\int\limits_{a}^{a}g(t) dt =0\]

Answer 20

http://puu.sh/7wGXz.png how are you able to do that?

Answer 21

and how does that approach the solution?

Answer 22

what do you mean how am I able to do that?

Answer 23

\[\int\limits\limits_{\frac{\pi}{2}}^{\pi} f(x) \cos(x) dx=f(x) \sin(x)|_\frac{\pi}{2}^\pi -\int\limits\limits_{\frac{\pi}{2}}^{\pi}\sin(x) \frac{e^x}{\sin(x)} dx \]
Plug in your limist for that first part...
\[=f(\pi)\sin(\pi)-f(\frac{\pi}{2})\sin(\frac{\pi}{2})-\int\limits_{\frac{\pi}{2}}^{\pi}\sin(x) \frac{e^x}{\sin(x)} dx\]

Answer 24

sin(pi)=?
f(pi/2)=?

Answer 25

this is what I've got so far http://puu.sh/7wHm3.png

Answer 26

then?

Answer 27

Don't forget your limits
also what is f'(x)? Use the fundamental thm of calculus

Answer 28

I get this http://puu.sh/7wHBy.png

Answer 29

for f'(x)

Answer 30

what next?

Answer 31

well i said don't forget the limits

Answer 32

how do you do that?

Answer 33

\[\int\limits\limits\limits_{\frac{\pi}{2}}^{\pi} f(x) \cos(x) dx=f(x) \sin(x)|_\frac{\pi}{2}^\pi -\int\limits\limits\limits_{\frac{\pi}{2}}^{\pi}\sin(x) \frac{e^x}{\sin(x)} dx \]

Answer 34

what answer do you get?

Answer 35

you will see when we finish simplifying this
i don't want to ruin the movie by giving away the last scene of it

Answer 36

please this is due in 10 minutes ;[

Answer 37

\[=f(\pi)\sin(\pi)-f(\frac{\pi}{2})\sin(\frac{\pi}{2})-\int\limits\limits_{\frac{\pi}{2}}^{\pi}\sin(x) \frac{e^x}{\sin(x)} dx \]
this is isn't even that long
you can do it
there is the next step

Answer 38

what is sin(pi)
?

Answer 39

0?

Answer 40

but how are you able to plug in values for f(x ) when you don't know the integral?

Answer 41

You have a definition of f(x):
\( \displaystyle f(x) = \int_{\pi/2}^{x} \frac{e^{t}}{\sin t} dt \)
The definition is valid when we choose a value for x such as when we want to take \(f (\frac{\pi}{2} ) \)
We simply substitute the value in for x directly into the definition.
If we did know the antiderivative of the integrand were a function A(t), then the function definition is equivalent to: f(x) = A(x) - A(pi/2). With x=pi/2, f(pi/2) = A(pi/2) - A(pi/2). Simply note that we have not changed anything other than x on both sides.

Answer 42

hmm

Answer 43

Oo this is a cool problem :)

Answer 44

example knore
pretend we look at f(x)=5x-2 from x=5 to x=5.
\[\int\limits_{5}^{5}f(x) dx=0\]

this is what the graph of the area between f(x)=5x-2 and y=0 from x=5 to x=5 looks like.

There is no area at all to look at.

Answer 45

I see what you mean now!

Answer 46

so no need to integrate at all!

Answer 47

Well yeah for that f function no integration needed

Answer 48

so the real value is (e^( pi )-e^(( pi )/(2)))

Answer 49

of the whole thing

Answer 50

well e^(pi/2)-e^pi

Answer 51

oh right the negative in front

Answer 52

Combining these results:
$$
\left (f(x)\sin(x)\right )'=f'(x)\sin(x)+cos(x)f(x)\\
\int\left (f(x)\sin(x)\right )'dx=\int \left (f'(x)\sin(x)+cos(x)f(x)\right ) dx\\
f(x)\sin(x=\int f'(x)\sin(x)dx+\int \cos(x)f(x) dx\\
\int f(x)\cos(x) dx=f(x)\sin(x)-\int f'(x)\sin(x)dx\\
$$
So,
$$
\large{
\int\limits\limits_{\frac{\pi}{2}}^{\pi} f(x) \cos(x) dx\\
=f(x) \sin(x)|_\frac{\pi}{2}^\pi -\int\limits\limits_{\frac{\pi}{2}}^{\pi}\sin(x) \frac{e^x}{\sin(x)} dx\\
=\large f(x) \sin(x)|_\frac{\pi}{2}^\pi - e^x|_\frac{\pi}{2}^\pi + c\\
=f(\pi)-e^{\pi}+e^{\pi/2}+c
}
$$
but
$$
\displaystyle f(x) = \int_{\pi/2}^{x} \frac{e^{t}}{\sin t} dt
$$
so
$$
f(\pi ) = \int_{\pi/2}^{\pi} \frac{e^{t}}{\sin t} dt
$$
Then
$$
\int\limits\limits_{\frac{\pi}{2}}^{\pi} f(x) \cos(x) dx\\
=\int_{\pi/2}^{\pi} \frac{e^{t}}{\sin t} dt-e^{\pi}+e^{\pi/2}+c
$$

However,

$$
\int_{\pi/2}^{\pi} \frac{e^{t}}{\sin t} dt
$$
is nonintegrable.

Answer 53

@ybarrap I think you have some error.

Answer 54

where @myininaya ?

Answer 55

"
\[=f(\pi)-e^{\pi}+e^{\pi/2}+c \]
"
first there shouldn't be plus c because we have a definite integral
second f(pi) is not suppose to be there

it should be
\[f(\pi) \sin(\pi)-f(\pi/2)\sin(\pi/2)-e^\pi+e^\frac{\pi}{2}\]
sin(pi)=0
f(pi/2)=0
so we have
-e^pi+e^pi/2

Answer 56

Oh, yeah. I see it! Thanks

Answer 57

I am a bit uncertain of something:
f(pi) sin (pi)
f(x) would diverge as we get to x=pi. (e^x/ sin x, sin x goes to 0). Would we need to be more careful of the indeterminate form sin (pi) * f(pi) "=" 0 * infinity ? It has been a while so I was just wondering what the justification is. :p

Answer 58

maybe that's irrelevant because sinpi is 0?

Answer 59

alright i looked it up for myself.

if it interests you, i just found a resource about lhopitals rule and figured it this way:
\( \displaystyle \lim_{x \to \pi-} \left( \sin x \int_{\pi/2}^{x} \frac{e^{t}}{\sin t} dt \right) \)
\( \displaystyle \lim_{x \to \pi-} \left( \frac{\int_{\pi/2}^{x} \frac{e^{t}}{\sin t} dt}{\csc x} \right) \)
\( \displaystyle \lim_{x \to \pi-} \left( \frac{\frac{d}{dx} \int_{\pi/2}^{x} \frac{e^{t}}{\sin t} dt}{\frac{d}{dx} \csc x} \right) \)
\( \displaystyle \lim_{x \to \pi-} \left( \frac{\frac{e^{x}}{\sin x}}{-\csc x \cot x} \right) \)
\( \displaystyle \lim_{x \to \pi-} \left( -e^{x} \frac{1}{\sin x} \frac{\sin x}{\cot x} \right) \)
\( \displaystyle \lim_{x \to \pi-} \left( -e^{x} \tan x \right) \) substitute directly.

Answer 60

@AccessDenied That was good observation on the whole improper integralness stuff.