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Question Number 212347 by Ismoiljon_008 last updated on 10/Oct/24

Commented by Ismoiljon_008 last updated on 10/Oct/24

$$$$$$\mathbb{H}elpme,please.$$$$$$

Answered by Ar Brandon last updated on 10/Oct/24

$$I={\int }_0^{\frac{\pi }{2}}{\left(\frac{\sqrt{\sin x}-\sqrt{\cos x}}{\sin x-\cos x}\right)}^{4}dx$$$$={\int }_0^{\frac{\pi }{2}}{\left(\frac{\sqrt{\sin x}-\sqrt{\cos x}}{(\sqrt{\sin x}-\sqrt{\cos x})(\sqrt{\sin x}+\sqrt{\cos x})}\right)}^{4}dx$$$$={\int }_0^{\frac{\pi }{2}}{\left(\frac{1}{\sqrt{\sin x}+\sqrt{\cos x}}\right)}^{4}dx={\int }_0^{\frac{\pi }{2}}{\left(\frac{\sqrt{\sec x}}{\sqrt{\tan x}+1}\right)}^{4}dx$$$$={\int }_0^{\frac{\pi }{2}}\frac{{\sec }^{2}x}{{(\sqrt{\tan x}+1)}^4}dx\stackrel{t=\tan x}{=}{\int }_0^{\mathrm{\infty }}\frac{dt}{{(\sqrt{t}+1)}^4}\stackrel{t=u^2}{=}2{\int }_0^{\mathrm{\infty }}\frac{u}{{(u+1)}^4}du$$$$=-{\left[\frac{2u}{3{(u+1)}^3}\right]}_0^{\mathrm{\infty }}+\frac{2}{3}{\int }_0^{\mathrm{\infty }}\frac{du}{{(u+1)}^3}=-{\left[\frac{1}{3{(u+1)}^2}\right]}_0^{\mathrm{\infty }}=\frac{1}{3}$$

Answered by Ghisom last updated on 10/Oct/24

$$\underset{0}{\stackrel{\pi /2}{\int }}{\left(\frac{\sqrt{\sin x}-\sqrt{\cos x}}{\sin x-\cos x}\right)}^{4}dx=$$$$[t=1+\sqrt{\tan x}]$$$$=2\underset{1}{\stackrel{\mathrm{\infty }}{\int }}(\frac{1}{t^3}-\frac{1}{t^4})dt={[\frac{2}{3t^3}-\frac{1}{t^2}]}_1^{\mathrm{\infty }}=\frac{1}{3}$$

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