In rectangular coordinates, the area is
[tex]\displaystyle\int_{-a}^a\int_{-(b/a)\sqrt{a^2-x^2}}^{(b/a)\sqrt{a^2-x^2}}\mathrm dy\,\mathrm dx[/tex]
or
[tex]\displaystyle\int_{-b}^b\int_{-(a/b)\sqrt{b^2-y^2}}^{(a/b)\sqrt{b^2-y^2}}\mathrm dx\,\mathrm dy[/tex]
We'll compute the first integral. By symmetry of the integrand, we have
[tex]\displaystyle\int_{-a}^a\int_{-(b/a)\sqrt{a^2-x^2}}^{(b/a)\sqrt{a^2-x^2}}\mathrm dy\,\mathrm dx=4\int_0^a\int_0^{(b/a)\sqrt{a^2-x^2}}\mathrm dy\,\mathrm dx[/tex]
The integral wrt [tex]y[/tex] is trivial, giving
[tex]\displaystyle\int_{-a}^a\int_{-(b/a)\sqrt{a^2-x^2}}^{(b/a)\sqrt{a^2-x^2}}\mathrm dy\,\mathrm dx=\frac{4b}a\int_0^a\sqrt{a^2-x^2}\,\mathrm dx[/tex]
In the remaining integral wrt [tex]x[/tex], substitute
[tex]x=a\sin t\implies\mathrm dx=a\cos t\,\mathrm dt[/tex]
to get
[tex]\displaystyle\int_{-a}^a\int_{-(b/a)\sqrt{a^2-x^2}}^{(b/a)\sqrt{a^2-x^2}}\mathrm dy\,\mathrm dx=4b\int_0^{\pi/2}\cos t\sqrt{a^2-(a\sin t)^2}\,\mathrm dt[/tex]
[tex]\displaystyle\int_{-a}^a\int_{-(b/a)\sqrt{a^2-x^2}}^{(b/a)\sqrt{a^2-x^2}}\mathrm dy\,\mathrm dx=4ab\int_0^{\pi/2}\cos t\sqrt{1-\sin^2t}\,\mathrm dt[/tex]
[tex]\displaystyle\int_{-a}^a\int_{-(b/a)\sqrt{a^2-x^2}}^{(b/a)\sqrt{a^2-x^2}}\mathrm dy\,\mathrm dx=4ab\int_0^{\pi/2}\cos t\sqrt{\cos^2t}\,\mathrm dt[/tex]
[tex]\displaystyle\int_{-a}^a\int_{-(b/a)\sqrt{a^2-x^2}}^{(b/a)\sqrt{a^2-x^2}}\mathrm dy\,\mathrm dx=4ab\int_0^{\pi/2}\cos t\,|\cos t|\,\mathrm dt[/tex]
Over the interval [tex]0<t<\dfrac\pi2[/tex], we have [tex]\cos t>0[/tex], so that [tex]|\cos t|=\cos t[/tex]. Then [tex]\cos t\,|\cos t|=\cos^2t[/tex], and recall a double angle identity, we get rewrite the integral as
[tex]\displaystyle\int_{-a}^a\int_{-(b/a)\sqrt{a^2-x^2}}^{(b/a)\sqrt{a^2-x^2}}\mathrm dy\,\mathrm dx=2ab\int_0^{\pi/2}(1+\cos2t)\,\mathrm dt[/tex]
The remaining integral is trivial.
[tex]\displaystyle\int_0^{\pi/2}(1+\cos2t)\,\mathrm dt=\left(t+\frac12\sin2t\right)\bigg|_{t=0}^{t=\pi/2}=\frac\pi2[/tex]
Then the area of the ellipse is
[tex]\displaystyle\int_{-a}^a\int_{-(b/a)\sqrt{a^2-x^2}}^{(b/a)\sqrt{a^2-x^2}}\mathrm dy\,\mathrm dx=\boxed{\pi ab}[/tex]
