4. int _(0)^pi /4cos(x)sqrt (2-sqrt (2)sin(x))dx

1. Substitution<br /> Let $u = \sqrt{2} \sin(x)$, then $du = \sqrt{2} \cos(x) dx$. Therefore, $\cos(x) dx = \frac{du}{\sqrt{2}}$.<br /><br />2. Change Limits<br /> When $x = 0$, $u = \sqrt{2} \sin(0) = 0$. When $x = \pi/4$, $u = \sqrt{2} \sin(\pi/4) = 1$.<br /><br />3. Transform Integral<br /> The integral becomes $\int_{0}^{1} \frac{1}{\sqrt{2}} \sqrt{2-u} \, du$.<br /><br />4. Simplify and Integrate<br /> Factor out $\frac{1}{\sqrt{2}}$: $\frac{1}{\sqrt{2}} \int_{0}^{1} \sqrt{2-u} \, du$. Use substitution $v = 2-u$, $dv = -du$. Limits change from $u=0$ to $v=2$ and $u=1$ to $v=1$. Thus, $\int_{2}^{1} \sqrt{v} (-dv) = \int_{1}^{2} \sqrt{v} \, dv$.<br /><br />5. Evaluate the Integral<br /> $\int_{1}^{2} v^{1/2} \, dv = \left[ \frac{2}{3} v^{3/2} \right]_{1}^{2} = \frac{2}{3} (2^{3/2} - 1^{3/2}) = \frac{2}{3} (2\sqrt{2} - 1)$.<br /><br />6. Final Calculation<br /> Multiply by $\frac{1}{\sqrt{2}}$: $\frac{1}{\sqrt{2}} \cdot \frac{2}{3} (2\sqrt{2} - 1) = \frac{2}{3} (\sqrt{2} - \frac{1}{\sqrt{2}}) = \frac{2}{3} (\sqrt{2} - \frac{\sqrt{2}}{2}) = \frac{2}{3} \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{3}$.