Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 70917 by Kunal12588 last updated on 09/Oct/19

$$\int \sqrt{{tan}^{2}x+3}dx$$

Commented by mathmax by abdo last updated on 09/Oct/19

$$\sqrt{3}t=tanx\Rightarrow x=arctan\left(t\sqrt{3}\right))\Rightarrow$$$$\int \sqrt{3+{tan}^{2}x}dx=\sqrt{3}\int \sqrt{1+t^2}\frac{\sqrt{3}}{1+3t^2}dt=3\int \frac{\sqrt{1+t^2}}{1+3t^2}dt$$$$changementt=sh\left(u\right)give$$$$\int \frac{\sqrt{1+t^2}}{1+3t^2}dt=\int \frac{ch\left(u\right)}{1+3{sh}^{2}u}chudu=\int \frac{\frac{1+ch\left(2u\right)}{2}}{1+3\frac{ch\left(2u\right)-1}{2}}du$$$$=\int \frac{1+ch\left(2u\right)}{2+3ch\left(2u\right)-3}du=\int \frac{ch\left(2u\right)+1}{3ch\left(2u\right)-1}du$$$$=\int \frac{\frac{e^{2u}+e^{-2u}}{2}+1}{3\frac{e^{2u}+e^{-2u}}{2}-1}du=\int \frac{e^{2u}+e^{-2u}+2}{3e^{2u}+3e^{-2u}-2}du$$$${=}_{e^{2u}=z}\int \frac{z+z^{-1}+2}{3z+3z^{-1}-2}\frac{dz}{2z}$$$$=\frac{1}{2}\int \frac{z+z^{-1}+2}{3z^2+3-2z}dz=\frac{1}{2}\int \frac{z^2+2z+1}{z(3z^2-2z+3)}dz$$$$letdecomposeF\left(z\right)=\frac{z^2+2z+1}{z(3z^2-2z+3)}$$$$3z^2-2z+3=0\to {\mathrm{\Delta }}^{{}^{\prime }}=1-9<0\Rightarrow F\left(z\right)=\frac{a}{z}+\frac{bz+c}{3z^2-2z+3}$$$$a=\frac{1}{3}$$$${lim}_{z\to +\mathrm{\infty }}zF\left(z\right)=\frac{1}{3}=a+\frac{b}{3}\Rightarrow 1=3a+b\Rightarrow b=1-3a=0\Rightarrow$$$$F\left(z\right)=\frac{1}{3z}+\frac{c}{3z^2-2z+3}$$$$F\left(1\right)=\frac{4}{4}=1=\frac{1}{3}+\frac{c}{4}\Rightarrow 4=\frac{4}{3}+c\Rightarrow c=4-\frac{4}{3}=\frac{8}{3}\Rightarrow$$$$F\left(z\right)=\frac{1}{3z}+\frac{8}{3(3z^2-2z+3)}\Rightarrow \frac{1}{2}\int F\left(z\right)dz=\frac{1}{6}ln\mid z\mid +\frac{8}{9}\int \frac{dz}{z^2-\frac{2}{3}z+1}$$$$\int \frac{dz}{z^2-\frac{2}{3}z+1}=\int \frac{dz}{z^2-2\frac{1}{3}z+\frac{1}{9}+1-\frac{1}{9}}$$$$=\int \frac{dz}{{(z-\frac{1}{3})}^2+\frac{8}{9}}{=}_{z-\frac{1}{3}=\frac{2\sqrt{2}}{3}u}\frac{9}{8}\int \frac{1}{1+u^2}\frac{2\sqrt{2}}{3}du$$$$=\frac{3\sqrt{2}}{4}arctan\left(\frac{3z-1}{2\sqrt{2}}\right)+c=\frac{3\sqrt{2}}{4}arctan\left(\frac{3e^{2u}-1}{2\sqrt{2}}\right)+cbut$$$$u=argsh\left(t\right)=ln(t+\sqrt{1+t^2})\Rightarrow 2u=2ln(t+\sqrt{1+t^2})$$$$\Rightarrow e^{2u}=(t+\sqrt{1+t^2})=(\frac{tanx}{\sqrt{3}}+\sqrt{1+\frac{{tan}^{2}x}{3}})\Rightarrow$$$$I=\frac{1}{3(\frac{1}{3}+\frac{2\sqrt{2}}{3}ln(\frac{tanx}{\sqrt{3}}+\sqrt{1+\frac{{tan}^{2}x}{3}})}+\frac{3\sqrt{2}}{4}arctan\left(\frac{3(\frac{tanx}{\sqrt{3}}+\sqrt{1+\frac{{tan}^{2}x}{3}}-1}{2\sqrt{2}}\right)$$$$+C.$$

Commented by Kunal12588 last updated on 10/Oct/19

$$thankyousir$$

Commented by mathmax by abdo last updated on 10/Oct/19

$$youarewelcome.$$

Answered by Kunal12588 last updated on 09/Oct/19

$$t=tanx$$$$\Rightarrow dx=\frac{dt}{1+t^2}$$$$I=\int \frac{\sqrt{t^2+3}}{1+t^2}dt$$$$now?$$$$oru=\sqrt{{tan}^{2}x+3}\Rightarrow dx=\frac{t}{(t^2-2)\sqrt{t^2-3}}dt$$$$I=\int \frac{t^2}{(t^2-2)\sqrt{t^2-3}}dt?...$$$$plshelp$$

Commented by MJS last updated on 09/Oct/19

$$\mathrm{easier}$$$$\int \frac{\sqrt{t^2+3}}{t^2+1}dt=$$$$[u=\frac{t}{\sqrt{t^2+3}}\to dt=\frac{\sqrt{{(t^2+3)}^3}}{3}]$$$$=-3\int \frac{du}{(u-1)(u+1)(2u^2+1)}=$$$$=-\frac{1}{2}\int \frac{du}{u-1}+\frac{1}{2}\int \frac{du}{u+1}+2\int \frac{du}{2u^2+1}=$$$$=-\frac{1}{2}\ln (u-1)+\frac{1}{2}\ln (u+1)+\sqrt{2}{\tan }^{-1}\left(\sqrt{2}u\right)=$$$$=\frac{1}{2}\ln \frac{u+1}{u-1}+\sqrt{2}{\tan }^{-1}\left(\sqrt{2}u\right)=...$$$$$$

Commented by MJS last updated on 09/Oct/19

$$\int \frac{\sqrt{t^2+3}}{t^2+1}dt=$$$$[u={\sinh }^{-1}\frac{t}{\sqrt{3}}\to dt=\sqrt{t^2+3}du]$$$$=3\int \frac{{\mathrm{e}}^{4u}+{2\mathrm{e}}^{2u}+1}{{3\mathrm{e}}^{4u}-{2\mathrm{e}}^{2u}+3}du=$$$$[v={\mathrm{e}}^{2u}\to du=\frac{dv}{{2\mathrm{e}}^{2u}}]$$$$=\frac{3}{2}\int \frac{{(v+1)}^2}{(3v^2-2v+3)v}dv=\frac{1}{2}\int \frac{dv}{v}+4\int \frac{dv}{3v^2-2v+3}=$$$$=\frac{1}{2}\ln v+\sqrt{2}{\tan }^{-1}\frac{\sqrt{2}(3v-1)}{4}=...$$

Commented by Kunal12588 last updated on 10/Oct/19

$$thnaksalotsir$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com