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Question Number 49367 by behi83417@gmail.com last updated on 06/Dec/18

$$a)\int \frac{\mathbf{dx}}{\sqrt{1-\mathbf{tgx}}}$$$$b)\int \frac{\mathbf{dx}}{\sqrt[3]{1-\mathbf{tgx}}}$$$$c)\int \frac{\mathbf{dx}}{\sqrt{1-\sqrt{1-\mathbf{x}}}}$$

Commented by behi83417@gmail.com last updated on 06/Dec/18

$$thankyouverymuchpro.Abdo.$$

Commented by maxmathsup by imad last updated on 06/Dec/18

$$youarewelcome.$$

Commented by maxmathsup by imad last updated on 06/Dec/18

$$c)trigonometricmethodchangementx={sin}^{2}tgive$$$$\int \frac{dx}{\sqrt{1-\sqrt{1-x}}}=\int \frac{2sintcost}{\sqrt{1-cost}}dt=2\int \frac{sintcost}{\sqrt{2}sin\left(\frac{t}{2}\right)}dt$$$$=2\int \frac{2sin\left(\frac{t}{2}\right)cos\left(\frac{t}{2}\right)}{\sqrt{2}sin\left(\frac{t}{2}\right)}(1-2{sin}^2\left(\frac{t}{2}\right))dt$$$$=\frac{4}{\sqrt{2}}\int (cos\left(\frac{t}{2}\right)-2cos\left(\frac{t}{2}\right){sin}^2\left(\frac{t}{2}\right))dt$$$$=\frac{4}{\sqrt{2}}\int cos\left(\frac{t}{2}\right)dt-\frac{8}{\sqrt{2}}\int cos\left(\frac{t}{2}\right){sin}^2\left(\frac{t}{2}\right)dt$$$$=\frac{8}{\sqrt{2}}sin\left(\frac{t}{2}\right)-\frac{16}{3\sqrt{2}}{sin}^3\left(\frac{t}{2}\right)+c$$$$=\frac{8}{\sqrt{2}}sin\left(\frac{arcsin\left(\sqrt{x}\right)}{2}\right)-\frac{16}{3\sqrt{2}}{sin}^3\left(\frac{arcsin\left(\sqrt{x}\right)}{2}\right)+c.$$

Commented by maxmathsup by imad last updated on 06/Dec/18

$$c)letI=\int \frac{dx}{\sqrt{1-\sqrt{1-x}}}changement\sqrt{1-x}=tgive1-x=t^{2}and$$$$I=\int \frac{-2tdt}{\sqrt{1-t}}=2\int \frac{1-t-1}{\sqrt{1-t}}dt=2\int \sqrt{1-t}dt-2\int \frac{dt}{\sqrt{1-t}}but$$$$\int \frac{dt}{\sqrt{1-t}}=-2\sqrt{1-t}and\int \sqrt{1-t}dt=\int {(1-t)}^{\frac{1}{2}}dt=-\frac{2}{3}{(1-t)}^{\frac{3}{2}}\Rightarrow$$$$I=-\frac{4}{3}{(1-t)}^{\frac{3}{2}}+4\sqrt{1-t}+c=-\frac{4}{3}{(1-\sqrt{1-x})}^{\frac{3}{2}}+4\sqrt{1-\sqrt{1-x}}+c.$$

Commented by behi83417@gmail.com last updated on 06/Dec/18

$$nicemethod.ilovethis.thanksalot.$$

Answered by MJS last updated on 06/Dec/18

$$\left(a\right)$$$$\int \frac{dx}{\sqrt{1-\tan x}}=$$$$[t=\sqrt{1-\tan x}\to dx=-\frac{2tdt}{t^4-2t^2+2}]$$$$=-2\int \frac{dt}{t^4-2t^2+2}=-2\int \frac{dt}{(t^2-\sqrt{2+2\sqrt{2}}t+\sqrt{2})(t^2+\sqrt{2+2\sqrt{2}}t+\sqrt{2})}=$$$$=-2\int \frac{dt}{(t^2-pt+q)(t^2+pt+q)}=$$$$=\frac{1}{pq}\int \frac{t-p}{t^2-pt+q}dt-\frac{1}{pq}\int \frac{t+p}{t^2+pt+q}dt$$$$\mathrm{and}\mathrm{these}\mathrm{can}\mathrm{be}\mathrm{solved}$$

Answered by tanmay.chaudhury50@gmail.com last updated on 06/Dec/18

$$a)\int \frac{dx}{\sqrt{1-tanx}}$$$$t^2=1-tanx2tdt=-{sec}^{2}xdx$$$$dx=\frac{-2tdt}{1+{(1-t^2)}^2}$$$$\int \frac{-2tdt}{t\times \{1+{(1-t^2)}^2\}}$$$$=-2\int \frac{dt}{t^4-2t^2+2}$$$$=-2\int \frac{\frac{1}{t^2}}{t^2-2+\frac{2}{t^2}}dt$$$$now{t}^2+\frac{2}{t^2}={(t+\frac{\sqrt{2}}{t})}^2-2\sqrt{2}$$$$={(t-\frac{\sqrt{2}}{t})}^2+2\sqrt{2}$$$$d(t+\frac{\sqrt{2}}{t})=(1-\frac{\sqrt{2}}{t^2})dt$$$$d(t-\frac{\sqrt{2}}{t})=(1+\frac{\sqrt{2}}{t^2})dt$$$$=\frac{-1}{\sqrt{2}}\int \frac{\frac{2\sqrt{2}}{t^2}}{t^2-2+\frac{2}{t^2}}dt$$$$=\frac{-1}{\sqrt{2}}\int \frac{(1+\frac{\sqrt{2}}{t^2})-(1-\frac{\sqrt{2}}{t^2})}{t^2-2+\frac{2}{t^2}}dt$$$$=\frac{-1}{\sqrt{2}}[\int \frac{1+\frac{\sqrt{2}}{t^2}}{{(t-\frac{\sqrt{2}}{t})}^2+2\sqrt{2}}dt-\int \frac{1-\frac{\sqrt{2}}{t^2}}{{(t+\frac{\sqrt{2}}{t})}^2-2\sqrt{2}}dt]$$$$=\frac{-1}{\sqrt{2}}[\frac{1}{\sqrt{2\sqrt{2}}}{tan}^{-1}\left(\frac{t-\frac{\sqrt{2}}{t}}{\sqrt{2\sqrt{2}}}\right)-\frac{1}{2\sqrt{2\sqrt{2}}}ln\left\{\frac{(t+\frac{\sqrt{2}}{t})-\sqrt{2\sqrt{2}}}{(t+\frac{\sqrt{2}}{t})+\sqrt{2\sqrt{2}}}\right\}+c$$$$nowplsputt=\sqrt{1-tanx}$$$$$$

Commented by behi83417@gmail.com last updated on 06/Dec/18

$$thanksinadvancesirMJSandsirtanmay.$$

Answered by tanmay.chaudhury50@gmail.com last updated on 06/Dec/18

$$c)t^2=1-\sqrt{1-x}$$$$(1-x)={(1-t^2)}^2$$$$1-x=1-2t^2+t^4$$$$x=t^4-2t^2$$$$dx=(4t^3-4t)dt$$$$\int \frac{4t^3-4t}{t}dt$$$$4\int t^{2}dt-4\int dt$$$$=\frac{4t^3}{3}-4t+c$$$$=\frac{4}{3}{\left\{(1-\sqrt{1-x})\right\}}^{\frac{3}{2}}-4{(1-\sqrt{1-x})}^{\frac{1}{2}}+c$$$$$$

Commented by tanmay.chaudhury50@gmail.com last updated on 06/Dec/18

$$tryingtosolveq.no\left(b\right)...excellentpostsfromyoursidesir..$$

Commented by behi83417@gmail.com last updated on 06/Dec/18

$$thankyousomuchsir.$$

Commented by behi83417@gmail.com last updated on 06/Dec/18

$$youarewellcomesir.oneofthebest$$$$questionsolverswithgreatworks$$$$inthislovelyforumis:sirtanmay.$$$$godlockdear.$$

Answered by MJS last updated on 06/Dec/18

$$\left(b\right)$$$$\int \frac{dx}{{(1-\tan x)}^{\frac{1}{3}}}=$$$$[t={(1-\tan x)}^{\frac{1}{3}}\to dx=-\frac{3t^2}{t^6-2t^3+2}]$$$$=-3\int \frac{t}{t^6-2t^3+2}dt$$$$t^6-2t^3+2=(t^2+\sqrt[3]{4}t+\sqrt[3]{2})(t^2-\frac{1-\sqrt{3}}{\sqrt[3]{2}}t+\sqrt[3]{2})(t^2-\frac{1+\sqrt{3}}{\sqrt[3]{2}}t+\sqrt[3]{2})=$$$$=(t^2+pt+q)(t^2+rt+q)(t^2+st+q)\Rightarrow$$$$\Rightarrow \frac{3}{q(p-r)(p-s)}\int \frac{t+p}{t^2+pt+q}dt+\frac{3}{q(r-p)(r-s)}\int \frac{t+r}{t^2+rt+q}dt+\frac{3}{q(s-p)(s-r)}\int \frac{t+s}{t^2+st+q}dt$$$$...\mathrm{and}\mathrm{again}\mathrm{we}\mathrm{can}\mathrm{solve}\mathrm{these}$$

Commented by behi83417@gmail.com last updated on 06/Dec/18

$$ok!thanks.waiting.....$$

Answered by tanmay.chaudhury50@gmail.com last updated on 07/Dec/18

$$b)t^3=1-tanx$$$$3t^{2}dt=-{sec}^{2}xdx$$$$dx=\frac{-3t^2}{1+{(1-t^3)}^2}dt=\frac{-3t^2}{t^6-2t^3+2}dt$$$$\int \frac{-3t^2}{t^6-2t^3+2}\times \frac{1}{t}dt$$$$=-3\int \frac{tdt}{t^6-2t^3+2}$$$$t^6-2t^3+2$$$$={\left(t^3\right)}^2-2\times t^3\times 1+1+1$$$$=-3\int \frac{tdt}{{(t^3-1)}^2-i^2}$$$$=-3\int \frac{tdt}{(t^3-1-i)(t^3-1+i)}$$$$=-3\int \frac{tdt}{(t^3+a)(t^3+b)}[a=-1-ib=-1+i]$$$$=\frac{-3}{a-b}\int \left[\frac{t(t^3+a)-t(t^3+b)}{(t^3+a)(t^3+b)}\right]dt$$$$=\frac{3}{b-a}[\int \frac{tdt}{t^3+b}-\int \frac{tdt}{t^3+a}]$$$$\mathit{n}\mathit{o}\mathit{w}\mathit{s}\mathit{o}\mathit{l}\mathit{v}\mathit{i}\mathit{n}\mathit{g}\int \frac{tdt}{t^3+a}$$$$\int \frac{tdt}{\left(t{+}^3\sqrt{a}\right)(t^2-t{\times }^3\sqrt{a}{+}^{\frac{2}{3}}\sqrt{a})}$$$$\int \frac{tdt}{(t+p)(t^2-tp+p^2)}$$$$\frac{t}{(t+p)(t^2-tp+p^2)}=\frac{A}{t+p}+\frac{Bt+C}{t^2-tp+p^2}$$$$t=A(t^2-tp+p^2)+(Bt+C)(t+p)$$$$t=t^2(A+B)+t(-Ap+Bp+C)+{Ap}^2+Cp$$$$A+B=0$$$${Ap}^2+Cp=0$$$$-Ap+Bp+C=1$$$$C=-Ap$$$$-Ap-Ap-Ap=1A=\frac{-1}{3p}B=\frac{1}{3p}C=\frac{-1}{3}$$$$\int \frac{A}{t+p}dt+\int \frac{Bt+C}{t^2-tp+p^2}dt$$$$\frac{-1}{3p}\int \frac{dt}{t+p}+\int \frac{\frac{1}{3p}t+\frac{-1}{3}}{t^2-tp+p^2}dt$$$$\frac{-1}{3p}\int \frac{dt}{t+p}+\frac{1}{3}\int \frac{t-1}{t^2-tp+p^2}dt$$$$\frac{-1}{3p}\int \frac{dt}{t+p}+\frac{1}{6}\int \frac{2t-p+p-2}{t^2-tp+p^2}$$$$\frac{-1}{3p}\int \frac{dt}{t+p}+\frac{1}{6}\int \frac{d(t^2-tp+p^2)}{t^2-tp+p^2}+\frac{p-2}{6}\int \frac{dt}{t^2-2.t.\frac{p}{2}+\frac{p^2}{4}+\frac{3p^2}{4}}$$$$\frac{-1}{3p}ln(t+p)+\frac{1}{6}ln(t^2-tp+p^2)+\frac{p-2}{6}\times \frac{1}{{\left(\frac{\sqrt{3}p}{2}\right)}^{}}\times {tan}^{-1}\left(\frac{t-\frac{p}{2}}{\frac{\sqrt{3}p}{2}}\right)+c_1$$$$\mathit{n}\mathit{o}\mathit{w}\mathit{p}\mathit{u}\mathit{t}\mathit{p}={\left(\mathit{a}\right)}^{\frac{1}{3}}$$$$\mathit{n}\mathit{e}\mathit{x}\mathit{t}\mathit{p}\mathit{u}\mathit{t}\mathit{a}=(-1-\mathit{i})$$$$\mathit{s}\mathit{i}\mathit{m}\mathit{i}\mathit{l}\mathit{a}\mathit{r}\mathit{l}\mathit{y}\mathit{w}\mathit{e}\mathit{c}\mathit{a}\mathit{n}\mathit{f}\mathit{i}\mathit{n}\mathit{d}\int \frac{tdt}{t^3+b}$$$$inplaceofpput{\left(b\right)}^{\frac{1}{3}}$$$$nextputb=(-1+i)$$$$finalyalgebricaddition...$$$$$$

Commented by behi83417@gmail.com last updated on 07/Dec/18

$$greatworkdonebyyou.thankyouverymuchsir.$$

Commented by tanmay.chaudhury50@gmail.com last updated on 07/Dec/18

$$thankyousir...$$

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