Support NeoGAF

[Suranga3]

Basically I can't figure out the antiderivative of the following: (note: { = intergral sign)
{(1 + sinx)^(1/2) dx

What I did so far is let u=1 + sinx, then du = cosx dx. I substituted du/cosx = dx back into the above formula to get.

{(u)^(1/2) * (du/cosx), and now I'm stuck. Please help. Much thanks.

 

[marsomega]

Basically I can't figure out the antiderivative of the following: (note: { = intergral sign)
{(1 + sinx)^(1/2) dx

What I did so far is let u=1 + sinx, then du = cosx dx. I substituted du/cosx = dx back into the above formula to get.

{(u)^(1/2) * (du/cosx), and now I'm stuck. Please help. Much thanks.

What methods of integration have you learned so far?

 

[Suranga3]

This is part of a question for a differential equations class. So, I'm assuming your allowed to use to any method of integration.

 

[Chrono]

I wanna know this too. >__<

 

[nitewulf]

well, with

int{u^(1/2)*du/cosx}, you could take the cosx outside the integral, as x isnt your variable of integration. with that,

(1/cox)*int{u^(1/2)*du} becomes a generic integral that could be integrated easily,

(1/cosx)*((1/(3/2))*u^(3/2))

ultimately simplifying to,

(2(1+sinx)^3/2)/(3cosx)

 

[cvxfreak]

Not having enough space in Trigonometry/Math Analysis this year was the best thing that ever happened to me.

 

[Chrono]

int{u^(1/2)*du/cosx}, you could take the cosx outside the integral, as x isnt your variable of integration. with that,

I thought you only can take out constants? Does cosx count as one here (it is a function of x right...? I mean I can't for example take out x^2 out of the integral...)?

;___;

 

[Chrono]

Ok I have this integration question... I'll be able to get the answer on monday so it's no big deal, but if I can know now i'll sleep better.

{ 1 / ( (lnx)^1/2 * x ) dx

That's the integral of one over the square root of lnx times x...

 

[nitewulf]

well the variable of integration is "u", and from the perspective of "u", x becomes a constant. like partial differentiation in multiple variables. say your function is,
f=3xy
if you wanted to differentiate it with respect to only "x",
f'=3y, y is seen as just a constant.
of course i did this stuff approximately 4 years ago and i might be forgetting something crucial.

 

[nitewulf]

ok thats a really easy one to get your feet wet with substitution.

u = lnx
du = (1/x)dx = dx/x

so,
int { dx/(xlnx)
int { (dx/x)*(1/lnx)
int { du/u = lnu = ln(lnx)

edit: sorry, missed the "1/2" power,

with that in mind, keeping everything the same,

int { du/u^(1/2)
int { u^(-1/2)du
and just use
1/(1+n) * u^(1+n) to get
2(lnx)^1/2

 

[Suranga3]

Thanks nitewulf!

 

[Chrono]

Thanks a lot.