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Question Number 53119 by rahul 19 last updated on 18/Jan/19

$$Evaluate:$$$$1)\int \sqrt{\frac{2-x}{4+x}}dx$$$$2)\int \sqrt{\frac{x-2}{x-4}}dx$$$$3)\int \sqrt{(x-2)(x-4)}dx$$$$4)\int \frac{dx}{2\sin \mathit{x}+3\sec \mathit{x}}.$$

Commented by maxmathsup by imad last updated on 18/Jan/19

$$3)letI=\int \sqrt{(x-2)(x-4)}dxchangement\sqrt{x-2}=tgivex-2=t^2\Rightarrow x=2+t^2$$$$I=\int t\sqrt{2+t^2-4}2tdt=2\int t^2\sqrt{t^2-2}dt{=}_{t=\sqrt{2}ch\left(u\right)}2\int 2{ch}^{2}t\sqrt{2}sh\left(u\right)\sqrt{2}sh\left(u\right)du$$$$=8\int {ch}^{2}u{sh}^{2}udu=8\int \frac{{\left(2shuchu\right)}^2}{4}du=2\int {sh}^{2}udu$$$$=2\int \frac{ch\left(2u\right)-1}{2}du=\int ch\left(2u\right)du-u+c$$$$=\frac{1}{2}sh\left(2u\right)-u+cbutu=argch\left(\frac{t}{\sqrt{2}}\right)=ln(\frac{t}{\sqrt{2}}+\sqrt{\frac{t^2}{2}-1})\Rightarrow$$$$sh\left(2u\right)=\frac{e^{2u}-e^{-2u}}{2}=\frac{1}{2}\{{(\frac{t}{\sqrt{2}}+\sqrt{\frac{t^2}{2}-1})}^2-{(\frac{t}{\sqrt{2}}+\sqrt{\frac{t^2}{2}-1})}^{-2}\}\Rightarrow$$$$I=\frac{1}{4}\{{(\frac{\sqrt{x-2}}{\sqrt{2}}+\sqrt{\frac{x-2}{2}-1})}^2-{(\frac{\sqrt{x-2}}{\sqrt{2}}+\sqrt{\frac{x-2}{2}-1})}^{-2}-ln(\frac{\sqrt{x-2}}{\sqrt{2}}+\sqrt{\frac{x-2}{2}-1})+c$$$$=\frac{1}{4}\{\frac{1}{2}{(\sqrt{x-2}+\sqrt{x-4})}^2-\frac{2}{{(\sqrt{x-2}+\sqrt{x-4})}^2}\}-ln\left(\frac{\sqrt{x-2}+\sqrt{x-4}}{\sqrt{2}}\right)+c$$

Commented by Tawa1 last updated on 18/Jan/19

$$\mathrm{Sir}\mathrm{please}\mathrm{check}\mathrm{your}\mathrm{solution}\mathrm{to}\mathrm{my}\mathrm{question}52841.$$$$\mathrm{i}\mathrm{asked}\mathrm{for}\mathrm{some}\mathrm{clarification}\mathrm{sir}.\mathrm{Thanks}\mathrm{sir}$$

Commented by maxmathsup by imad last updated on 23/Jan/19

$$2)letI=\int \sqrt{\frac{x-2}{x-4}}dxchangement\sqrt{\frac{x-2}{x-4}}=tgive\frac{x-2}{x-4}=t^2\Rightarrow$$$$x-2=t^{2}x-4t^2\Rightarrow (1-t^2)x=2-4t^2\Rightarrow x=\frac{-4t^2+2}{1-t^2}=\frac{4t^2-2}{t^2-1}\Rightarrow$$$$\frac{dx}{dt}=\frac{8t(t^2-1)-(4t^2-2)2t}{{(t^2-1)}^2}=\frac{8t^3-8t-8t^3+4t}{{(t^2-1)}^2}=\frac{-4t}{{(t^2-1)}^2}\Rightarrow$$$$I=\int t\left(\frac{-4t}{{(t^2-1)}^2}\right)dt=-4\int \frac{t^2}{{(t^2-1)}^2}dt$$$$=-4\int \frac{t^2-1+1}{{(t^2-1)}^2}dt=-4\int \frac{dt}{t^2-1}-4\int \frac{dt}{{(t^2-1)}^2}but$$$$\int \frac{dt}{t^2-1}=\frac{1}{2}\int (\frac{1}{t-1}-\frac{1}{t+1})dt=\frac{1}{2}ln\mid \frac{t-1}{t+1}\mid +c_1$$$$letdecomposeF\left(t\right)=\frac{1}{{(t^2-1)}^2}\Rightarrow F\left(t\right)=\frac{1}{{(t-1)}^2{(t+1)}^2}$$$$=\frac{a}{t-1}+\frac{b}{{(t-1)}^2}+\frac{c}{(t+1)}+\frac{d}{{(t+1)}^2}$$$$b={lim}_{t\to 1}{(t-1)}^{2}F\left(t\right)=\frac{1}{4}$$$$d={lim}_{t\to -1}{(t+1)}^{2}F\left(t\right)=\frac{1}{4}\Rightarrow F\left(t\right)=\frac{a}{t-1}+\frac{1}{4{(t-1)}^2}+\frac{c}{t+1}+\frac{1}{4{(t+1)}^2}$$$$F(-t)=F\left(t\right)\Rightarrow \frac{-a}{t+1}+\frac{1}{4{(t+1)}^2}+\frac{-c}{t-1}+\frac{1}{4{(t-1)}^2}=F\left(t\right)\Rightarrow c=-a\Rightarrow$$$$F\left(t\right)=\frac{a}{t-1}+\frac{1}{4{(t-1)}^2}-\frac{a}{t+1}+\frac{1}{4{(t+1)}^2}$$$$F\left(0\right)=1=-a+\frac{1}{4}-a+\frac{1}{4}\Rightarrow -2a+\frac{1}{2}=1\Rightarrow -2a=\frac{1}{2}\Rightarrow a=-\frac{1}{4}\Rightarrow$$$$F\left(x\right)=-\frac{1}{4(t-1)}+\frac{1}{4(t+1)}+\frac{1}{4{(t-1)}^2}+\frac{1}{4{(t+1)}^2}\Rightarrow$$$$\int \frac{dt}{{(t^2-1)}^2}=\frac{1}{4}ln\mid \frac{t+1}{t-1}\mid -\frac{1}{4(t-1)}-\frac{1}{4(t+1)}+c_2\Rightarrow$$$$I=-2ln\mid \frac{t-1}{t+1}\mid -ln\mid \frac{t+1}{t-1}\mid +\frac{1}{t-1}+\frac{1}{t+1}+C\Rightarrow$$$$I=ln\mid \frac{t+1}{t-1}\mid +\frac{1}{t-1}+\frac{1}{t+1}+C$$$$I=ln\mid \frac{\sqrt{\frac{x-2}{x-4}}+1}{\sqrt{\frac{x-2}{x-4}}-1}\mid +\frac{1}{\sqrt{\frac{x-2}{x-4}}-1}+\frac{1}{\sqrt{\frac{x-2}{x-4}}+1}+C.$$

Answered by MJS last updated on 18/Jan/19

$$\left(1\right)$$$$t=\sqrt{\frac{4+x}{2-x}}\to x=\frac{2(t^2-2)}{t^2+1};dx=\frac{\sqrt{{(2-x)}^3(4+x)}}{3}dt$$$$12\int \frac{dt}{{(t^2+1)}^2}$$$$u=\arctan t\to t=\tan u;dt=(t^2+1)du$$$$12\int {\cos }^{2}udu$$

Answered by MJS last updated on 18/Jan/19

$$\left(2\right)$$$$t=\sqrt{\frac{x-4}{x-2}}\to x=\frac{2(t^2-2)}{(t^2-1)};dx=\sqrt{(x-4){(x-2)}^3}dt$$$$4\int \frac{dt}{{(t^2-1)}^2}$$$$u=\arcsin \frac{1}{t}\to t=\frac{1}{\sin u};dt=-t\sqrt{t^2-1}du$$$$-4\int \frac{{\sin }^{2}u}{{\cos }^{3}u}du=-4\int {\sec }^{3}u-\sec udu$$

Answered by MJS last updated on 18/Jan/19

$$\left(3\right)$$$$t=x-3\to dt=dx$$$$\int \sqrt{t^2-1}dt$$$$u=\arccos \frac{1}{t}\to t=\frac{1}{\cos u};dt=t\sqrt{t^2-1}du$$$$\int \frac{{\sin }^{2}u}{{\cos }^{3}u}du...$$

Answered by tanmay.chaudhury50@gmail.com last updated on 18/Jan/19

$$1)\int \frac{2-x}{\sqrt{(4+x)(2-x)}}dx$$$$\int \frac{2-x}{\sqrt{8-4x+2x-x^2}}dx$$$$\int \frac{2-x}{\sqrt{8-2x-x^2}}$$$$\frac{1}{2}\int \frac{-2-2x+6}{\sqrt{8-2x-x^2}}dx$$$$=\frac{1}{2}\int \frac{d(8-2x-x^2)}{\sqrt{8-2x-x^2}}+3\int \frac{dx}{\sqrt{{\left(3\right)}^2-{(x+1)}^2}}$$$$=\frac{1}{2}\times \frac{{(8-2x-x^2)}^{\frac{-1}{2}+1}}{\frac{-1}{2}+1}+3{sin}^{-1}\left(\frac{x+1}{3}\right)+c$$$$={(8-2x-x^2)}^{\frac{1}{2}}+3{sin}^{-1}\left(\frac{x+1}{3}\right)+c$$$$2)\int \frac{x-2}{\sqrt{(x-2)(x-4)}}dx$$$$\int \frac{x-2}{\sqrt{x^2-6x+8}}dx$$$$\frac{1}{2}\int \frac{2x-6+2}{\sqrt{x^2-6x+8}}dx$$$$\frac{1}{2}\int \frac{d(x^2-6x+8)}{\sqrt{x^2-6x+8}}+\int \frac{dx}{\sqrt{{(x-3)}^2-1}}$$$$\frac{1}{2}\times \frac{{(x^2-6x+8)}^{\frac{-1}{2}+1}}{\frac{-1}{2}+1}+ln\{(x-3)+\sqrt{{(x-3)}^2-1}\}+c$$$$={(x^2-6x+8)}^{\frac{1}{2}}+ln\{(x-3)+\sqrt{x^2-6x+8}\}+c$$$$3\int \sqrt{(x-2)(x-4)}dx$$$$\int \sqrt{{(x-3)}^2-1}dx$$$$=\frac{(x-3)}{2}\sqrt{{(x-3)}^2-1}-\frac{1}{2}ln\{(x-3)+\sqrt{{(x-3)}^2-1}\}+c$$$$=\frac{(x-3)}{2}\sqrt{x^2-6x+8}-\frac{1}{2}ln\{(x-3)+\sqrt{(x^2-6x+8}$$$$$$

Commented by rahul 19 last updated on 18/Jan/19

$$ThankyouSir!$$

Commented by Otchere Abdullai last updated on 18/Jan/19

$$wow!welldonesir!$$

Answered by MJS last updated on 18/Jan/19

$$\left(4\right)$$$$=\int \frac{\cos x}{3+\cos x\sin x}dx$$$$\mathrm{Weyerstrass}$$$$t=\tan \frac{x}{2}\to x=2\arctan t;dx={2\cos }^2\frac{x}{2}dt=\frac{2dt}{t^2+1}$$$$\cos x=-\frac{t^2-1}{t^2+1};\sin x=\frac{2t}{t^2+1}$$$$-2\int \frac{t^2-1}{(t^2-2t+3)(3t^2+2t+1)}dt$$$$\mathrm{now}\mathrm{decompose}...$$$$\frac{t^2-1}{(t^2-2t+3)(3t^2+2t+1)}=\frac{1}{8}\times \frac{t+1}{t^2-2t+3}-\frac{3}{8}\times \frac{t+1}{3t^2+2t+1}$$$$\frac{t+1}{t^2-2t+3}=\frac{t-1}{t^2-2t+3}+\frac{2}{t^2-2t+3}$$$$\frac{t+1}{3t^2+2t+1}=\frac{6t+2}{6(3t^2+2t+1)}+\frac{4}{6(3t^2+2t+1)}=$$$$=\frac{1}{3}\times \frac{3t+1}{3t^2+2t+1}+\frac{2}{3}\times \frac{1}{3t^2+2t+1}$$$$\mathrm{now}\mathrm{use}\mathrm{formulas}$$

Commented by rahul 19 last updated on 18/Jan/19

$$ThankYouSir!$$

Answered by tanmay.chaudhury50@gmail.com last updated on 18/Jan/19

$$4)\int \frac{dx}{2sinx+3secx}$$$$\int \frac{cosxdx}{3+2sinxcosx}$$$$\frac{1}{2}\int \frac{cosx-sinx+cosx+sinx}{3+2sinxcosx}dx$$$$\frac{1}{2}\int \frac{cosx-sinx}{2+{(sinx+cosx)}^2}dx+\frac{1}{2}\int \frac{d(sinx-cosx)}{4-(1-2sinxcosx)}$$$$\frac{1}{2}\int \frac{d(sinx+cosx)}{{\left(\sqrt{2}\right)}^2+{(sinx+cosx)}^2}+\frac{1}{2}\int \frac{d(sinx-cosx)}{4-{(sinx-cosx)}^2}\leftarrow formula\int \frac{dx}{a^2-x^2}$$$$\frac{1}{2}\times \frac{1}{\sqrt{2}}{tan}^{-1}\left(\frac{sinx+cosx}{\sqrt{2}}\right)+\frac{1}{2}\times \frac{1}{2\times 2}ln\left(\frac{2+sinx-cosx}{2-sinx+cosx}\right)+c$$

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