1. ## Integral Question

Umm, I'm not quite sure how to ask this....

$\displaystyle \int {x^3 e^{x^2 } } dx = \tfrac{1} {2}\int {x^2 e^{x^2 } } d(x^2 )$

How do you do that? What's going on here? As in changing dx to d(x^2). Is this a method I haven't been taught?....or do I not fully understand the concept of "with respect to x"? I see that d(x^2) equals 2x giving you the same integral but I need further clarification....

Thanks.

2. This is just an alternative notation for making a substitution in an integral. If you replace $\displaystyle x^2$ by $\displaystyle y$ in the right-hand integral then it becomes $\displaystyle \tfrac{1}{2}\!\!\int y e^y\,dy$, which exactly what you would get by making the substitution $\displaystyle y=x^2$ in the left-hand integral.

3. Originally Posted by RedBarchetta
Umm, I'm not quite sure how to ask this....

$\displaystyle \int {x^3 e^{x^2 } } dx = \tfrac{1} {2}\int {x^2 e^{x^2 } } d(x^2 )$

How do you do that? What's going on here? As in changing dx to d(x^2). Is this a method I haven't been taught?....or do I not fully understand the concept of "with respect to x"? I see that d(x^2) equals 2x giving you the same integral but I need further clarification....

Thanks.
$\displaystyle d(x^2) = 2 x \, dx$. Substitute and you get your original integral.

Perhaps a more obvious approach for you is to make the substitution

$\displaystyle u = x^2 \Rightarrow \frac{du}{dx} = 2x \Rightarrow dx = \frac{du}{2x}$

$\displaystyle \frac{1}{2} \int u e^u \, du$ which can be solved using integration by parts.