Thread: Is thins transformation valid ?

1. Is thins transformation valid ?

Okay , i have the Integral of xdx/[(x-2.5)^2-6.25] Substitution |t=x-2.5;dt=dx;x=t+2.5|
= Integral of tdt/(t^2-6.25) + 2.5 * Integral of dt/(t^2-6.25) ; So i can solve the first one with substitution , and the second part is where my question comes out.
Can i return the substitution of t back as 2.5 * Integral of dx/[(x-2.5)^2-6.25]

Seems pretty fine to me, but had to be sure if im not making any mistake.

2. Re: Is thins transformation valid ?

Writing this "pretty", you want to integrate $\int \frac{x dx}{(x- 2.5)^2- 6.25}$. (I presume you "completed the square" to get that denominator.)

Yes, you can make the substitution u= x- 2.5 so that du= dx and x= u+ 2.5. That makes the integral $\int \frac{(u+ 2.5)du}{u^2- 6.25}= \int \frac{u du}{u^2- 6.25}- 2.5\int \frac{du}{u^2- 6.25}$

For the second of those integrals you are asking "Can i return the substitution of t back as 2.5 * Integral of dx/[(x-2.5)^2-6.25]?"

You could but there is no point! It is simpler to do as is:
$2.5 \int \frac{du}{u^2- 6.25}= 2.5\int \frac{du}{(u- 2.5)(u+ 2.5)}$
Use "partial fractions".

3. Re: Is thins transformation valid ?

Yea i didnt thought of that too.. Actually now i see i wouldnt make any change (facepalm).. At first i used a trigonometric substitution as in your example u=sqrt(6.25)secZ ; du= sqrt(6.25)secZtanZdZ ; but that drove me to a more complicated integrals + because this is a definite integral im solving i forgot that to use the substitution u=sqrt(6.25)secZ , there is condition u>=a which in my case wasnt. So i returned backwards and the first thing i though was mentioned in my question.

Anyway thanks again, i feel like i still don`t have the sense of some quick and easy transformations in integrals, so sometimes i must do the hard way first .

4. Re: Is thins transformation valid ?

But just saying, I always find things like S dx/(x^2-a^2) very sketchy since you can get coth^-1 or tanh^-1 depending on the domain...