Can't do this integral

I honestly haven't got the slightest clue. I just recognised the form from another question I'd seen.



Do you mean co-ordinates or ordinates? They are two different (but somewhat linked) terms.

At first glance, it seems like you need to rearrange your first equation for $t$ in terms of $x$, substitute it into your second equation so that you have $y$ in terms of $x$ and finally use your point to find the value of $a$.

Co-ordinates, sorry.

Where do you get all these fiendish questions from?

I had a go and made some progress by taking the 2nd equation and writing it in exponential form. By going down that route you can solve the 2nd equation for t and make progress from there.

My physics teacher and the internet

what do you mean, writing it in exponential form?

I don't know if this is right, but my approach was to take this equation

t^2 * sint = pi^2 / 4 and write sine in exponential format

i.e change this equation to

i/2*t^2(exp(it) - exp(-it)) = pi^2 / 4

and go from there. I mean t has to be real, so thats a useful property I would be looking to exploit.

Guys halp me again pls pls

$\, \int \dfrac {sinx}{sinx-1} dx$

Guys halp me again pls pls

$\, \int \dfrac {sinx}{sinx-1} dx$

Well-spotted.

Thanks

I just finished the integral with that sub and it yielded a pretty cool result.

$\displaystyle \begin{aligned} \int \dfrac{\sin x}{\sin x - 1} \text{ d}x & \overset{t=\tan \frac{x}{2}}= \ x - \dfrac{2}{1 - \tan \frac{x}{2}} + \mathcal{C} \\ & \ \ \equiv \ x - \dfrac{2\cos \frac{x}{2}}{\cos \frac{x}{2} - \sin \frac{x}{2}} + \mathcal{C} \end{aligned}$

The world's sneakiest substitution!

Isn't it just Also, I had a little read about integral transforms. Laplace is such a boss!$\mathcal{L} \left[ f(t) \right](s) = \displaystyle \int_{0}^{\infty} f(t) e^{-ts} \text{ d}t$

All about Fourier man

Back with one more:

$\int \dfrac {1}{1+\sqrt{x}} dx$

PS how do I the big integral sign?

$u=\sqrt{x}$

To do a big integral sign you need to add "\displaystyle" before the "\int" command.

Just did it lol, the exact same substitution. Well, I used $x=u^2$ but that's the same thing lol.

And thanks!

Moar:

$\displaystyle \int sin^x(\dfrac{\pi}{6})$

$\displaystyle\int \sin^x\left(\dfrac{\pi}{6}\right)\, \mathrm{d} x=\int 2^{-x}\, \mathrm{d} x = -\frac{2^{-x}}{\log{2}}+C$

Where are you getting these from?