How can I find the integral of this expression: $$\int_{-a}^{a}\frac{x*dy}{( x^2 + y^2 )^{\frac{3}{2}}}$$
Please would yo people give me a step by step solution. I am stuck in this
Finding an Integral of $\int_{-a}^{a}\frac{x*dy}{( x^2 + y^2 )^{\frac{3}{2}}}$
-4
$\begingroup$
integration
definite-integrals
-
0i think there is no elementary solution – 2017-02-17
-
0Since the integral is with respect to $dy$, do you mean that $x$ is simply a constant (and thus can be factored out)? Or is there something else, some other relation, that you didn't mention? – 2017-02-17
-
11. factor out the $x's$ and substitute $u=y/x$ so that you see that you just need to find $\int du/(1+u^2)^{3/4}.$ 2. Plug that into mathematica and see that the answer is in terms of hypergeometric functions. 3. Ask yourself 'have I written the problem down correctly?' – 2017-02-17
-
1I am sorry people I made an error in the definition of the integral. Now I have just made an edit. Yes @zipirovich x is just a constant. – 2017-02-17
-
1thank you @spaceisddarkgreen ! Yes there was a mistake in the problem I wrote! sorry – 2017-02-17
1 Answers
0
As suggested, we have
$$\begin{align} \int_{y=-a}^{a} \frac{x\,dy}{(x^{2}+y^{2})^{\frac{3}{2}}} &\,=\, \frac{1}{x^{2}}\int_{y=-a}^{a} \frac{dy}{(1+\left(\frac{y}{x}\right)^{2})^{\frac{3}{2}}} \,=\, \frac{1}{x}\int_{u=-\frac{a}{x}}^{\frac{a}{x}} \frac{du}{(1+u^{2})^{\frac{3}{2}}} \\[0.3cm] &\,=\, \left.\frac{1}{x}\cdot \frac{u}{\sqrt{u^{2}+1}}\right|_{u=-\frac{a}{x}}^{\frac{a}{x}} \end{align}$$
using the substitution $y=ux$. You should be able to take it from there.
-
1Thank you so much. I did not see this would do the trick – 2017-02-17