Definite Integration Test - 3

This integral is done over the unit square region in the first quadrant:

Convert this to polar co-ordinates:

$\displaystyle \int_{0}^{1} \int_{0}^{1} \frac{dxdy}{ \left(1+x^{2} +y^{2}\right)^{\frac{3}{2}}}=\displaystyle \int_{0}^{\frac{\pi}{4}} \int_{0}^{\sec \theta} \frac{r dr d\theta}{ \left(1+r^{2}\right)^{\frac{3}{2}}} +\displaystyle \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \int_{0}^{\csc \theta} \frac{r dr d\theta}{ \left(1+r^{2}\right)^{\frac{3}{2}}}$ . The bounds are this because from 0 to $\frac{\pi}{4}$ , the distance from origin to the side of the square will be the $r$ variable and projection of this $r$

to x-axis will always be 1, giving $r\cos\theta =1, r=\sec\theta$ . the second bounds are similarly justified

we can do each one separately and sum them. For first one: $\displaystyle \int_{0}^{\frac{\pi}{4}} \int_{0}^{\sec \theta} \frac{r dr d\theta}{ \left(1+r^{2}\right)^{\frac{3}{2}}} =\displaystyle \int_{0}^{\frac{\pi}{4}} d\theta-\frac{d\theta}{\sqrt{1+\sec^{2} \theta}}=\displaystyle \int_{0}^{\frac{\pi}{4}} d\theta-\frac{\cos\theta d\theta}{\sqrt{1+\cos^{2} \theta}}=\displaystyle \int_{0}^{\frac{\pi}{4}} d\theta-\frac{\cos\theta d\theta}{\sqrt{2-\sin^{2} \theta}}$ , this standard integral can be done by simple substitution of $u=\frac{\sin \theta}{\sqrt2}$ , and the indefinite integral is $\theta -\arcsin\left(\frac{\sin \theta}{\sqrt 2}\right)$ .

The second part of the integral can also be done in similar manner and summing them after evaluating them at the bounds will give $\frac{\pi}{6}$

Let us integrate with respect to $y$ first. For this, let us assume $1+x^2=z^2$ . Within the given limits, the value of the integral is $\dfrac{1}{z^2\sqrt {z^2+1}}$ . Then substitute for $z$ , and integrate with respect to $x$ . Again within the given limits, the value of the integral is $\tan^{-1} \displaystyle {(\dfrac{1}{√3})=\dfrac{π}{6}}$ . So, $a=1, b=6$ , making $a+b=\boxed {7}$