Question

The value $\frac{{\int }_0^{\pi /2}{\left(\sin x\right)}^{\sqrt{2}+1}dx}{{\int }_0^{\pi /2}{\left(\sin x\right)}^{\sqrt{2}-1}dx}$ is

• A. $\frac{\sqrt{2}+1}{\sqrt{2}-1}$
• B. $\frac{\sqrt{2}-1}{\sqrt{2}+1}$
• C. $\frac{\sqrt{2}+1}{\sqrt{2}}$
• D. $2-\sqrt{2}$

Answer 1

The correct option is D $2-\sqrt{2}$

$I_1={\int }_0^{\frac{\pi }{2}}{\sin }^{\sqrt{2}+1}xdx={\int }_0^{\frac{\pi }{2}}{\sin }^{\sqrt{2}}x.\sin xdx{I}_2={\int }_0^{\frac{\pi }{2}}{\sin }^{\sqrt{2}-1}xdx$

Applying integration by parts on $I_1$

$I_1={\left(\right({\sin x)}^{\sqrt{2}}\int \sin xdx)}_0^{\frac{\pi }{2}}-{\int }_0^{\frac{\pi }{2}}(\sqrt{2}{\sin }^{\sqrt{2}-1}x\cos x\int \sin xdx)dx{I}_1=-{(\cos x{\sin }^{\sqrt{2}}x)}_0^{\frac{\pi }{2}}-{\int }_0^{\frac{\pi }{2}}(\sqrt{2}{\sin }^{\sqrt{2}-1}x\cos x(-cosx)I_1=-{(\cos x{\sin }^{\sqrt{2}}x)}_0^{\frac{\pi }{2}}+{\int }_0^{\frac{\pi }{2}}(\sqrt{2}{\sin }^{\sqrt{2}-1}x{\cos }^{2}x)dx{I}_1=-{(\cos x{\sin }^{\sqrt{2}}x)}_0^{\frac{\pi }{2}}+\sqrt{2}{\int }_0^{\frac{\pi }{2}}{\sin }^{\sqrt{2}-1}x(1-{\sin }^{2}x)dx{I}_1=-{(\cos x{\sin }^{\sqrt{2}}x)}_0^{\frac{\pi }{2}}+\sqrt{2}({\int }_0^{\frac{\pi }{2}}{\sin }^{\sqrt{2}-1}x+{\int }_0^{\frac{\pi }{2}}{\sin }^{\sqrt{2}+1}x{I}_1=0+\sqrt{2}(I_2+I_1\left)\right(1-\sqrt{2})I_1=\sqrt{2}{I}_2\frac{I_1}{I_2}=\frac{\sqrt{2}}{(1-\sqrt{2})}=\frac{\sqrt{2}(1-\sqrt{2})}{(1-\sqrt{2})(1-\sqrt{2)}}=2-\sqrt{2}$

So option $D$ is correct