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

$$\int^{\pi}_{5\pi/6}\frac{\cos^4x}{\sqrt{1-\sin x}}dx=\frac{117\sqrt3}{140\sqrt2}-\frac{18}{35}$$
$$A=\int^{\pi}_{5\pi/6}\frac{\cos^4x}{\sqrt{1-\sin x}}dx$$ Multiply both numerator and denominator with $\sqrt{1+\sin x}$ $$A=\int^{\pi}_{5\pi/6}\frac{\cos^4x}{\sqrt{1-\sin x}}\frac{\sqrt{1+\sin x}}{\sqrt{1+\sin x}}dx$$ $$A=\int^{\pi}_{5\pi/6}\frac{\cos^4x\sqrt{1+\sin x}}{\sqrt{1-\sin^2x}}dx$$ $$A=\int^{\pi}_{5\pi/6}\frac{\cos^4x\sqrt{1+\sin x}}{\sqrt{\cos^2x}}dx$$ $$A=\int^{\pi}_{5\pi/6}\frac{\cos^4x\sqrt{1+\sin x}}{-\cos x}dx$$ (because on $[5\pi/6,\pi]$, $\cos x\lt0$) $$A=-\int^{\pi}_{5\pi/6}\cos^3 x\sqrt{1+\sin x}dx$$ $$A=\int^{5\pi/6}_{\pi}\cos^3 x\sqrt{1+\sin x}dx$$ We can rewrite $$\cos^3x=\cos^2x\cos x=(1-\sin^2x)\cos x=(1-\sin x)(1+\sin x)\cos x$$ $$A=\int^{5\pi/6}_{\pi}(1-\sin x)(1+\sin x)\sqrt{1+\sin x}\cos xdx$$ Then set $u=1+\sin x$, so $\sin x=u-1$ and $1-\sin x=1-(u-1)=2-u$ Also, $du=\cos xdx$ - For $x=\pi$, $u=1+\sin\pi=1$ - For $x=5\pi/6$, $u=1+\sin(5\pi/6)=1+1/2=3/2$ $$A=\int^{3/2}_1(2-u)u\sqrt udu=\int^{3/2}_1(2-u)u^{3/2}du$$ $$A=\int^{3/2}_1(2u^{3/2}-u^{5/2})du$$ $$A=\Big(2\times\frac{2u^{5/2}}{5}-\frac{2u^{7/2}}{7}\Big)\Big]^{3/2}_1$$ $$A=\Big(\frac{4}{5}(\frac{3}{2})^{5/2}-\frac{2}{7}(\frac{3}{2})^{7/2}\Big)-\Big(\frac{4}{5}-\frac{2}{7}\Big)$$ $$A=\Big(\frac{4}{5}\frac{9\sqrt3}{4\sqrt2}-\frac{2}{7}\frac{27\sqrt3}{8\sqrt2}\Big)-\frac{18}{35}$$ $$A=\frac{9\sqrt3}{5\sqrt2}-\frac{27\sqrt3}{28\sqrt2}-\frac{18}{35}$$ $$A=\frac{117\sqrt3}{140\sqrt2}-\frac{18}{35}$$