## Thomas' Calculus 13th Edition

$y=\int_{-1}^1(\sqrt {2-sin^2u})du +2=2$
Step 1. Letting $f(x)=\sqrt {2-sin^2x}$ and $F(x)=\int_a^x f(u)du+C$, we have $F'(x)=f(x)$ where $C$ is a constant. Step 2. In the case of the exercise, we can write $y=\int_a^x f(u)du +C$, where $C$ is a constant. Step 3. Given $y(-1)=2$, we have $y(-1)=\int_a^{-1} f(u)du +C=2$; thus we can let $a=-1, C=2$ Step 4. The solution can be written as $y=\int_{-1}^{1} f(u)du +2$