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