Two questions regarding integrating items

[TFM]

Homework Statement

I have a couple of questions about some intriguing looking equations.

Firstly, I have to integrate:

\frac{sin\theta}{cos\theta} d\theta

Is there an answer to just dividing them? because sin/cos = tan, but that doesn't seem to help?

Secondly, I have to integrate:

6e^{-x^2}
This One > 6e^{-x^2} Should be displaying e to the minus x sqaured
but I didn't think you could integrate a e^{x^2}

Again, this one V
e^{x^2} ?

Homework Equations

\frac{sin\theta}{cos\theta} = tan\theta

The Attempt at a Solution

Given above

Any help would be most appreciated,

TFM

 

[rock.freak667]

\int \frac{sin\theta}{cos\theta}d\theta


if you let u=cos\theta, what is \frac{du}{d\theta}?

For the 2nd one, I don't think you can express that in terms of elementary functions.

 

[TFM]

\int \frac{sin\theta}{cos\theta}d\theta


if you let u=cos\theta, what is \frac{du}{d\theta}?

\frac{du}{d\theta} = -sin\theta

For the 2nd one, I don't think you can express that in terms of elementary functions.

Is there any way to integrate this, or should I leav it as it is (The question is solving differential equations)

TFM

 

[rock.freak667]

\frac{du}{d\theta} = -sin\theta

and so -du=sin\theta d\theta.

Now just sub this into the integral and you can get it out.

 

[TFM]

I have to admit making a small mistake...

The equation to solve is:

\frac{cos\theta}{tan\theta}

TFM

 

[rock.freak667]

then you need to put tan as sin/cos and get

\frac{cos^2 \theta}{sin\theta}

which in this case put u=sin

 

[TFM]

So:

u = sin\theta

\frac{du}{d\theta} = cos\theta

du = cos\theta d\theta

so sub-ing in:

\frac{cos^2 \theta}{sin\theta}d\theta

\frac{du^2 \theta}{u}d\theta

Does this look right?

Edit: No it doesn't. Should be

\frac{du^2 }{u}

\frac{du^2 }{u}
Does this Look right?

TFM

 

[Gib Z]

I think what rock.freak667 wanted you to get was

\frac{du}{u} \cdot \cos x = \frac{ \sqrt{1-u^2}}{u} du,

which would warrant an integration by parts.

An alternative method would be to write \cos^2 x = 1-\sin^2 x and divide through.

 

[TFM]

Could I not use double angle formula:

cos^2(\theta) = \frac{1-cos(2\theta)}{2}

Giving

\frac{1 - cos(2\theta)}{2sin(\theta)}

and then let u = cos\theta again

??

TFM

 

[HallsofIvy]

How would that help you? What are you going to do with the cos(2\theta) if u= cos(\theta)?

 

[TFM]

That's true, I just saw the cos squared and remebered that sometime that relation is used to break it down. So:

\frac{cos^2 \theta}{sin\theta}

u = sin\theta

\frac{du}{d\theta} = cos\theta

du = cos\theta d\theta

This gives me

\frac{cos\theta cos\theta}{sin\theta}d\theta

\frac{cos\theta }{u}du


cos\theta = \sqrt{1 - sin^2\theta}

cos\theta = \sqrt{1 - u^2}

so:


\frac{\sqrt{1 - u^2}}{u}du

So integrating by parts:

\int adb = ab - \int bda

db = u^{-1} ; u^{-2}

a = \sqrt{1 - u^2} ; da =

Does this look okay?

TFM

 

[Gib Z]

Well if you meant

db = u^{-1}, b = \log_e u

Then that looks fine so far in terms of correctness, though it looks like it won't be fruitful. Perhaps rock.freak667 meant something different to what i Interpreted.

Why didn't you like my suggestion though lol?

 

[TFM]

Well if you meant

db = u^{-1}, b = \log_e u

Then that looks fine so far in terms of correctness, though it looks like it won't be fruitful. Perhaps rock.freak667 meant something different to what I Interpreted.

Why didn't you like my suggestion though lol?

I was actually going to do

b = u^{-1}, db = u^{-2}

or is this going the wrong way?

Was your suggestion


An alternative method would be to write cos^2 x = 1-sin^2 x and divide through.

If so, should that not be

cos^2(\theta) = \frac{1-cos(2\theta)}{2} ?

I'm wondering if i have made some mistakes in actually solving the Differential Equations, since out of four DEs I have three which seem to not be able to be completed?

TFM

 

[TFM]

If I may just check my calculations.

First Question:

x(t): x' = xt + 6te^{-t^{2}}

Subtract -xt from both sides:

-xt \frac{dx}{dt} = 6te^{-t^{2}}

-x \frac{dx}{dt} = 6te^{-t^{2}}/t

-x dx = 6e^{-t^{2}} dt

Does this one look okay?

TFM

 

[TFM]

Does the above look correct?

TFM

 

[rock.freak667]

\frac{dx}{dt}=xt+6te^{-t^2}

\frac{dx}{dt}-xt=6te^{-t^2}

Do you know how to solve first order differential equations of the form (in your case)

\frac{dx}{dt}+P(t)x=Q(t)?

 

[TFM]

That would be:

= e^Fx \int^F q(t)dt + C

= e^Fx \int^F q(t)dt + C

= e^Fx \int^F q(t)dt + C

F = \int p(t)dt

?

TFM

 

[TFM]

Sorry, Latex annoys me sometimes when editing:

= e^Fx \int^F q(t)dt + C

F = \int p(t)dt

TFM

 

[TFM]

Lets see. So:

\frac{dx}{dt}-xt=6te^{-t^2}

\frac{dx}{dt}+P(t)x=Q(t)

solution:

= e^Fx \int^F q(t)dt + C

F = \int p(t)dt

And:

q(t) = 6te^{-t^2}

p(t) = -t

so:

F = \int -t dt = \frac{-t^2}{2}

= e^{\frac{-t^2}{2}}x \int^{\frac{-t^2}{2}}(6te^{-t^2}) dt + C

We still seem to have to integrate the e^{-x^2}? Which I thought ewas impossible?

TFM

 

[rock.freak667]

your integrating factor of exp(-t²/2) is correct.

But your integrand should be


e^\frac{t^2}{2} 6te^{t^2}

Which can be simplified. Then you can integrating using a u substitution.

 

[TFM]

with

e^\frac{t^2}{2} 6te^{t^2}

can you add the e values to get:

6te^{\fract{t^2}{2}+t^2}

?

TFM

 

[rock.freak667]

exp(-t²/2 - t²)=exp(-3t²/2)

 

[TFM]

So we have:

e^{\frac{\t^2}{2}}\int 6te^{\frac{-3t^2}{2}}

So we have to integrate

\int 6te^{\frac{-3t^2}{2}}

You suggest using a u substitution

Should that be:

u = \frac{-3t^2}{2}

?

TFM

 

[rock.freak667]

Yes, if u=exp(-3t²/2), What is du/dt and thus du?

 

[TFM]

So:

U = e^{-\frac{3t^2}{2}}

Would du/dt be:

\frac{du}{dt} = \frac{3}{2}^{-\frac{3t^2}{2}}

and thus:

du = \frac{3}{2}^{-\frac{3t^2}{2}} dt

?

TFM