Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 158781 by HongKing last updated on 08/Nov/21

Answered by MJS_new last updated on 09/Nov/21

$$\underset{0}{\stackrel{\pi }{\int }}\frac{(1+{\sin }^{2}2x)\cos 4x}{1+a{\sin }^{2}x}dx=2\underset{0}{\stackrel{\pi /2}{\int }}\frac{(1+{\sin }^{2}2x)\cos 4x}{1+a{\sin }^{2}x}$$$$[t=\tan x\to dx=\frac{dt}{t^2+1}]$$$$=2\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{t^8-34t^4+1}{{(t^2+1)}^4((a+1)t^2+1)}dx=$$$$\left[{\mathrm{Ostrogradski}}^{\prime }\mathrm{s}\mathrm{Method}\right]$$$$=-{\left[\frac{8t(3(a+1)(a-4)t^4+(7a^2-24a-24)t^2-3(5a+4))}{3a^3{(t^2+1)}^3}\right]}_0^{\mathrm{\infty }}+$$$$-\frac{8(a+2)(a^2-4a-4)}{a^4}\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{dt}{t^2+1}+$$$$+\frac{2(a^2-4a-4)(a^2+8a+8)}{a^4}\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{dt}{(a+1)t^2+1}=$$$$=-{\left[\frac{8t(3(a+1)(a-4)t^4+(7a^2-24a-24)t^2-3(5a+4))}{3a^3{(t^2+1)}^3}\right]}_0^{\mathrm{\infty }}-$$$$-{\left[\frac{8(a+2)(a^2-4a-4)}{a^4}\arctan t\right]}_0^{\mathrm{\infty }}+$$$$+{\left[\frac{2(a^2-4a-4)(a^2+8a+8)}{a^4\sqrt{a+1}}\arctan \left(\sqrt{a+1}t\right)\right]}_0^{\mathrm{\infty }}=$$$$=\frac{(a^2-4a-4)(a^2+8a+8-4(a+2)\sqrt{a+1})}{a^4\sqrt{a+1}}\pi$$$$$$$$\frac{\pi }{a}+\frac{(a^2-4a-4)(a^2+8a+8-4(a+2)\sqrt{a+1})}{a^4\sqrt{a+1}}\pi =0$$$$\Rightarrow$$$$(3a^3-8a^2-48a-32)\sqrt{a+1}=(a^2-4a-4)(a^2+8a+8)$$$$\mathrm{squaring}\mathrm{\&}\mathrm{transforming}$$$$a^3(a^5-a^4-a^3-80a^2-144a-64)=0$$$$a\ne 0\Rightarrow \mathrm{answer}\mathrm{is}2$$$$\mathrm{but}\mathrm{still}\mathrm{the}\mathrm{only}\mathrm{solution}\mathrm{for}a\mathrm{is}<0$$$$a\approx -.938890161126$$

Commented by HongKing last updated on 09/Nov/21

$$\mathrm{thank}\mathrm{you}\mathrm{so}\mathrm{much}\mathrm{my}\mathrm{dear}\mathbf{S}\mathrm{er},$$$$\mathrm{but}\mathrm{its}\mathrm{not}\mathrm{correct},\mathrm{check}\mathrm{from}\mathrm{the}$$$$6\mathrm{th}\mathrm{line}..$$

Commented by HongKing last updated on 09/Nov/21

$$\mathrm{my}\mathrm{dear}\mathrm{Ser},\mathrm{thank}\mathrm{you}\mathrm{so}\mathrm{much}\mathrm{for}$$$$\mathrm{the}\mathrm{awesome}\mathrm{solution}$$

Commented by MJS_new last updated on 09/Nov/21

$$\mathrm{corrected}\mathrm{it}$$

Answered by ajfour last updated on 09/Nov/21

$$I={\int }_0^{\pi /2}\frac{(3-\cos 4x)\cos 4x}{1+\frac{a}{2}(1-\cos 2x)}dx$$$$=-\frac{2\pi }{a}$$$$I={\int }_0^{\pi /2}\frac{(3-\cos 4x)\cos 4x}{1+\frac{a}{2}(1+\cos 2x)}dx$$$$2I={\int }_0^{\pi /2}\frac{(2+a)(3-\cos 4x)\cos 4x}{{(1+\frac{a}{2})}^2-\frac{a^2}{8}(1+\cos 4x)}dx$$$$\frac{a^{2}I}{8(2+a)}=$$$${\int }_0^{\pi /4}\frac{(4-s-1)(s+1-1)}{s+1-2{(\frac{2}{a}+1)}^2}dx=\frac{\pi }{4(\frac{2}{a}+1)}$$$$says+1=1+\cos 4x=w$$$${\int }_0^{\pi /4}\frac{(4-w)(w-1)}{w-2k^2}dx=\frac{\pi }{4k}$$$$=-\int (w+2k^2)dx+\int \frac{4k^2-4+5w}{w-2k^2}dx$$$$=-(1+2k^2)\frac{\pi }{4}+\frac{5\pi }{4}+\int \frac{14k^2-4}{2\cos {}^{2}2x-2k^2}dx$$$$=\pi -\frac{k^2\pi }{2}+(\frac{1}{k^2}-\frac{7}{2}){\int }_0^{\mathrm{\infty }}\frac{dt}{t^2+1-\frac{1}{k^2}}$$$$=\frac{\pi }{2}(2-k^2)+\left(\frac{2-7k^2}{2k^2}\right)\frac{k}{\sqrt{k^2-1}}\left(\frac{\pi }{2}\right)$$$$=\frac{\pi }{4k}$$$$\Rightarrow 4k-2k^3+\frac{2-7k^2}{\sqrt{k^2-1}}=1$$$$\Rightarrow$$$$(k^2-1){(2k^3-4k+1)}^2={(2-7k^2)}^2$$$$k\approx 2.33581ork\approx -2.42122$$$$As1+\frac{2}{a}=k$$$$a\approx \frac{2}{k-1}buta>0\Rightarrow k>1$$$$hencea\approx 1.49722$$$$--------------$$$$butwehavetofindQ.$$$$Q^6=a(a^4-a^3-a^2-80a-144)$$

Commented by HongKing last updated on 10/Nov/21

$$\mathrm{thank}\mathrm{you}\mathrm{so}\mathrm{much}\mathrm{my}\mathrm{dear}\mathrm{Ser}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com