Question

prove integral

Answer 1

\[\int\limits_{0}^{\pi} xf(sinx)dx= \frac{ \pi }{ 2 }\int\limits_{0}^{\pi}f(sinx)dx\]

Answer 2

use substitution u=pi-x

Answer 3

its quite frustrating..

Answer 4

i get to: \[2\int\limits_{0}^{\pi} xf(sinx)dx=0\] and i'm stuck

Answer 5

\[\sin (\pi-x)=\sin x\\t=\pi-x\]
\[-\int _\pi^0(\pi-t)f(\sin t)dt=\int_0^\pi\pi f(\sin t)-\int_0^\pi t f(\sin t)=\int_0^\pi t f(\sin t)\]
\[\pi \int_0^\pi f(\sin t)=2\int_0^\pi t f(\sin t)\]
\[\frac{\pi}{2}\int_0^\pi f(\sin t)=\int_0^\pi t f(\sin t)\]

Answer 6

why use: \[−\int\limits_{\pi}^{0}(π−t)f(sint)dt\]

Answer 7

@Jonask

Answer 8

and t=pi-x, so how is this correct

Answer 9

its a substitution
notice that \[dx=-dt\] also
\[x=\pi-t\] ,if \[x=0,t=\pi\\if \\x=\pi ,t=0\]substitute this into
\[\int _0^\pi x f(\sin x)dx\]

Answer 10

i will check this tomorrow again. but i still don't understand how \[−\int\limits\limits_{\pi}^{0}(π−t)f(sint)dt\] helps you

Answer 11

\[\int _0^\pi xf(\sin x)dx=\int _\pi^0(\pi-t)f(\sin t)(-dt)=-\int _\pi^0(\pi-t)f(\sin t)dt\]

Answer 12

jus say what is it that challenges u here,that wich u dnt undrstanf

Answer 13

i get: \[\int\limits_{0}^{\pi}xf(sinx)dx=\int\limits_{0}^{\pi}(\pi-t)f(sint)dt\]

Answer 14

@Jonask

Answer 15

thats where i'm stuck

Answer 16

@Jonask help please

Answer 17

t=pi-x
dt = -dx

do the limits now :-
lower limit : x = 0, => t = pi-0 = pi
upper limit : x = pi, => t = pi - pi = 0

Answer 18

done

Answer 19

\[\int\limits\limits_{0}^{\pi}xf(sinx)dx=-\int\limits\limits_{\pi}^{0}(\pi-t)f(sint)dt\]
<=> \[\int\limits\limits\limits_{0}^{\pi}xf(sinx)dx=\int\limits\limits\limits_{0}^{\pi}(\pi-t)f(sint)dt\]

Answer 20

<=> \[\int\limits\limits\limits_{0}^{\pi}xf(sinx)dx=\pi\int\limits\limits\limits_{0}^{\pi}(f(sint)dt-\int\limits\limits\limits_{0}^{\pi}(tf(sint)dt\]

Answer 21

am i doing this right?

Answer 22

yes so far so good

Answer 23

then this is where something goes wrong

Answer 24

add the second integral both sides

Answer 25

\(\int\limits\limits\limits_{0}^{\pi}xf(sinx)dx=\pi\int\limits\limits\limits_{0}^{\pi}(f(sint)dt-\int\limits\limits\limits_{0}^{\pi}(tf(sint)dt \)

add \(\int\limits\limits\limits_{0}^{\pi}(tf(sint)dt\) both sides

Answer 26

\[\int\limits\limits\limits\limits_{0}^{\pi}xf(sinx)dx+\int\limits\limits\limits\limits_{0}^{\pi}tf(sint)dt=\pi\int\limits\limits\limits\limits_{0}^{\pi}f(sint)dt\]

Answer 27

and x is not equal to t, right?

Answer 28

Yes, next we use this :-
for a definite integral, variable of integration doesnt matter. variable is just dummy. f(x) is same as f(t)

Answer 29

where can i find this property?

Answer 30

\(\int\limits\limits\limits_{0}^{\pi}(tf(sint)dt = \int\limits\limits\limits_{0}^{\pi}(xf(sinx)dx \)

Answer 31

usually any course on definite integrals starts wid this property;
we can make sense of this property easily. lets see :)

Answer 32

let me put it this way:-

\(\int\limits\limits\limits_{0}^{\pi}(tf(sint)dt =F(\pi) - F(0)\)

clearly the dummy variable \(t\) is NOT playing any role here. it just disappears in the final result.

Answer 33

see if that satisfies you hmm

Answer 34

hmmm...

Answer 35

nope, don't get it

Answer 36

i'll check my textbook for this property

Answer 37

Also check the attached snapshot from my textbook also :)

Answer 38

oh nice, thanks!

Answer 39

closed

Answer 40

cool :)

Answer 41

thanks again

Answer 42

np.. u wlc :)