Question

Integration question #2

Answer 1

How should I start?

Answer 2

Take out your pencil/pen. For a start.

Answer 3

lol ishaan

Answer 4

@Ishaan94 Oh... Thank you....

Answer 5

remember the COH- cos=0pp./hyp.

Answer 6

\[I_2 =\int_0^{\pi}g(\pi-x) \ln(1 + e^{\cos(\pi - x)}) \]

Answer 7

\[2I = \int_0^{\pi} -g(x)\ln(1 + e^{-\cos x}) + g(x)\ln(1+e^{\cos x})\]
\[2I = \int _)^{\pi} g(x) \left(\ln \frac{1 + e^{\cos x}}{1 + \frac1{e^{\cos x}}}\right)\]

Answer 8

\[I = \frac12 \int_0^{\pi}g(x)\cos x\]

Answer 9

Watch this video, and only then you will become a JEDI Knight of Math.

http://www.youtube.com/watch?v=vEanZ-Ex5Uw

hehe ...kidding.

Answer 10

If you don't understand which property @Ishaan94 used, Here it is

\[\int\limits_{a}^{b}f(x)dx = \int\limits_{a}^{b}f(a+b-x)dx\]

Answer 11

Yes. Yes.

Answer 12

Actually... In the question... It stated that using the substitution u=pi -x ...

Answer 13

It's the same. You can try it on your own.

Answer 14

Oh!~ Finally I got it! Thanks !!!

Answer 15

AWWW i missed it :/ is there a #3???

Answer 16

I understand Ishaan's answer. But I can't do it with u-sub:S
Yea.. perhaps there is a #3

Answer 17

ohhh NOOOOOOOOO, i lost my answer due to stupid google chrome crash.

@callisto perform the substitution, I will guide you. I don't wanna type the answer again. Do the substitution and then tell me what you get.

Answer 18

NICE ONE @Ishaan94 :))

Answer 19

\[I = \int_{0}^{\pi}g(x)\ln (1+e^{cosx})dx\]\[I = -\int_{0}^{\pi}g(\pi-x)\ln (1+e^{cos(\pi-x)})dx\]
Let u = π - x
du = -dx , when x=0, u=π ; when x=π, u=0\[I = \int_{\pi}^{0}g(u)\ln (1+e^{cos(u)})du\]
Up to this step, any problem?

Answer 20

oh thanks :-)

Answer 21

-cosu not cosu

Answer 22

wait...

Answer 23

yes negative cosu

Answer 24

Why?

Answer 25

cos(pi - x) =/= cos x but cos(pi - x) = -cos x

Answer 26

Hmm... if it is not u-sub I understand this concept. So, when we do u-sub, we need to change the sign also?

Answer 27

I meant if we need to consider the quad. it lies

Answer 28

of course.

Answer 29

waait... why did you change the limit?

Answer 30

u-sub....Shouldn't I change the limit?

Answer 31

Ignore all my earlier advice and follow this.
\[I=\int _0^{\pi} g(\pi - x) \ln (1 +e^{\cos(\pi-x)})dx\]After the U substitution\[I=-\int_0^{\pi}g(u)\ln(1+e^{\cos u})du\tag1\]From (1)
\[I =-\int_0^{\pi} g(\pi -u)\ln (1+e^{\cos(\pi - u)})\implies I =\int_0^{\pi} g(u)\ln (1+e^{-\cos u})\tag2 \]
Now add 1 and 2.

Answer 32

So, still we should do u-sub first. Then, replace x by u in the condition given ( or vice versa). Next, do something sensible, right?

Answer 33

yeah you will have to replace the u with x.

Answer 34

Okay, that makes sense now. Thank you. And what's wrong with your previous solution?

Answer 35

Nothing is wrong with my previous solution? (The Latex one)

Answer 36

Alright.... Now I understand the u-sub, not your previous one :S But thanks!