does anyone have a clue how to integrate e^(sin x) ? I'm totally stumped..

Simple

If I am not wrong should be:

1/(cos x) e^(sin x)

If I am not wrong should be:

1/(cos x) e^(sin x)

i assume you mean: sec(x)e^(sin x) + C

but if you differentiate that, using the product and chain rules, then you get: sec(x)tan(x)e^(sin(x)) + sec(x)cos(x)e^(sin(x))

also using the trapezium method:

but

so that can't be it.

but if you differentiate that, using the product and chain rules, then you get: sec(x)tan(x)e^(sin(x)) + sec(x)cos(x)e^(sin(x))

also using the trapezium method:

```
```

?

?

? e^(sin x) dx =~ 6.175

?

0

but

```
```

? ??

? sec(x)e^(sin x) ? = (-2)

? ?0

so that can't be it.

I'll give it a shot :grin:

try this algo, I cant be botheret to write all I just wroted

1. substition sinx=t

then it will be:

dt=cosx*dx

dx=dt/sqrt(1-(sinx)^2) => dx=dt/sqrt(1-t^2)

2. you will get

I = Integral( e^t/sqrt(1-t^2)) dx

partial integration:

u = e^t ; du=e^t *dt

dv=dt/sqrt(1-t^2) ; v = arcsin(x)

3. you will get

I = arcsin(x) *e^t - Integral( arcsin(t) *e^t)dt

I = A - B

B = Integral( arcsin(t) *e^t)dt

partial again:

e^t=u ; du=e^t *dt

arcsin(x)*dt=dv ; v=........ find it

I didnt finished it but I think now it should hopefully comes down to an easier integral,

if not well you cant blame man for trying :)

do you know that joke about derivate and functions, ps joke for math geek only :)

One functions are walking down the street and suddenly it saws all other functions

running away like hell. It stops one of them and asks,

function A - hey whats going on, why are everybody running away?

function B - everyone is running becouse we heard that derivate comming here... you should be gone too

function A - naaah screw the derivate, I am e^x

try this algo, I cant be botheret to write all I just wroted

1. substition sinx=t

then it will be:

dt=cosx*dx

dx=dt/sqrt(1-(sinx)^2) => dx=dt/sqrt(1-t^2)

2. you will get

I = Integral( e^t/sqrt(1-t^2)) dx

partial integration:

u = e^t ; du=e^t *dt

dv=dt/sqrt(1-t^2) ; v = arcsin(x)

3. you will get

I = arcsin(x) *e^t - Integral( arcsin(t) *e^t)dt

I = A - B

B = Integral( arcsin(t) *e^t)dt

partial again:

e^t=u ; du=e^t *dt

arcsin(x)*dt=dv ; v=........ find it

I didnt finished it but I think now it should hopefully comes down to an easier integral,

if not well you cant blame man for trying :)

do you know that joke about derivate and functions, ps joke for math geek only :)

One functions are walking down the street and suddenly it saws all other functions

running away like hell. It stops one of them and asks,

function A - hey whats going on, why are everybody running away?

function B - everyone is running becouse we heard that derivate comming here... you should be gone too

function A - naaah screw the derivate, I am e^x

Here it is:

We open e^x to Maclaurin series and integrate term by term. It is a long way though.

e^x = 1 + x + x^2/2 + x^3/3! + ... + x^n/n! ...

f(x) = e^sin(x) = 1 + sin(x) + sin^2(x)/2 + sin^3(x)/6 + sin^4(x)/24 + ...

int f(x) dx = x - cos(x) - sin(2x)/4 + x/2 - sin^2(x)cos(x)/3 - 2cos(x)/3 ... + C

By the way

int sin^m(x) dx = -sin^(m-1)(x) cos(x)/m + (m-1)int(sin^(m-2))dx /m

We open e^x to Maclaurin series and integrate term by term. It is a long way though.

e^x = 1 + x + x^2/2 + x^3/3! + ... + x^n/n! ...

f(x) = e^sin(x) = 1 + sin(x) + sin^2(x)/2 + sin^3(x)/6 + sin^4(x)/24 + ...

int f(x) dx = x - cos(x) - sin(2x)/4 + x/2 - sin^2(x)cos(x)/3 - 2cos(x)/3 ... + C

By the way

int sin^m(x) dx = -sin^(m-1)(x) cos(x)/m + (m-1)int(sin^(m-2))dx /m

thanks guys :alright: i never thought of doing a series. nice joke btw mikky :D i think your method does work if you keep on integrating, but inFinie your method works better. thx again both :)

Yeah, a nice joke... except you forgot the punchline :P Oh well. I guess it is a nice test to see if your favourite math geek is drunk.