Evaluate the integral by making an appropriate change of variables. 2 sin(25x2 + 49y2) da, r where r is the region in the first quadrant bounded by the ellipse 25x2 + 49y2 = 1

[tex]25x^2+49y^2=1[/tex]

Notice that if

[tex]x=\dfrac15\cos t[/tex]
[tex]y=\dfrac17\sin t[/tex]

then

[tex]\implies25\left(\dfrac15\cos t\right)^2+49\left(\dfrac17\sin t\right)^2=1[/tex]
[tex]\implies \cos^2t+\sin^2t=1[/tex]

which suggests a change of variables of

[tex]\begin{cases}x(r,t)=\dfrac r5\cos t\\\\y(r,t)\dfrac r7\sin t\end{cases}[/tex]

where [tex]0\le r\le1[/tex] and [tex]0\le t\le2\pi[/tex]. Then we also have

[tex]25x^2+49y^2=r^2[/tex]

So

[tex]\displaystyle2\iint_{\mathcal R}\sin(25x^2+49y^2)\,\mathrm dA=2\int_0^{2\pi}\int_0^1\sin(r^2)\,r\,\mathrm dr\,\mathrm dt=4\pi\int_0^1r\sin(r^2)\,\mathrm dr[/tex]

Now a replacement of [tex]r\to r^2[/tex] will finish this. We get

[tex]\displaystyle2\pi\int_0^1(2r)\sin(r^2)\,\mathrm dr=2\pi\int_0^1\sin(r^2)\,\mathrm d(r^2)[/tex]
[tex]=-2\pi\cos(r^2)\bigg|_{r^2=0}^{r^2=1}=-2\pi(\cos(1)-\cos(0))=2\pi(1-\cos(1))[/tex]