## University Calculus: Early Transcendentals (3rd Edition)

$$\int^{\pi/6}_{0}\sqrt{1+\sin x}dx=2-\sqrt2$$
$$A=\int^{\pi/6}_{0}\sqrt{1+\sin x}dx$$ Multiply both numerator and denominator with $\sqrt{1-\sin x}$ $$A=\int^{\pi/6}_{0}\sqrt{1+\sin x}\times\frac{\sqrt{1-\sin x}}{\sqrt{1-\sin x}}dx$$ $$A=\int^{\pi/6}_{0}\frac{\sqrt{1-\sin^2x}}{\sqrt{1-\sin x}}dx$$ $$A=\int^{\pi/6}_{0}\frac{\sqrt{\cos^2x}}{\sqrt{1-\sin x}}dx$$ $$A=\int^{\pi/6}_{0}\frac{\cos x}{\sqrt{1-\sin x}}dx$$ (because on $[0,\pi/6]$, $\cos x\gt0$) We set $u=\sqrt{1-\sin x}$, which means $$du=\frac{-\cos x}{2\sqrt{1-\sin x}}dx$$ $$\frac{\cos x}{\sqrt{1-\sin x}}dx=-2du$$ - For $x=\pi/6$, $u=\sqrt{1-1/2}=\sqrt{1/2}$ - For $x=0$, $u=\sqrt{1-0}=1$ $$A=-2\int^{\sqrt{1/2}}_1du=2\int^1_{\sqrt{1/2}}du$$ $$A=2u\Big]^1_{\sqrt{1/2}}$$ $$A=2\Big(1-\sqrt{\frac{1}{2}}\Big)=2\Big(1-\frac{\sqrt2}{2}\Big)$$ $$A=2-\sqrt2$$