Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 111818 by bemath last updated on 05/Sep/20

$$\sqrt{bemath}$$$$\left(1\right)\int \frac{\cos x}{2-\cos x}dx$$$$\left(2\right)f\left(x\right)=\mid x^3\mid \Rightarrow f{}^{\prime }\left(x\right)?$$

Answered by bobhans last updated on 05/Sep/20

$$\mathrm{f}\left(\mathrm{x}\right)=\mid {\mathrm{x}}^3\mid \to \{\begin{array}{l}{\mathrm{x}}^3;\mathrm{x}>0\\ 0;\mathrm{x}=0\\ -{\mathrm{x}}^3;\mathrm{x}<0\end{array}$$$$\mathrm{f}{}^{\prime }\left(\mathrm{x}\right)=\{\begin{array}{l}{3\mathrm{x}}^2;\mathrm{x}>0\\ 0;\mathrm{x}=0\\ -{3\mathrm{x}}^2;\mathrm{x}<0\end{array}$$$$\mathrm{f}{}^{\prime }\left(\mathrm{x}\right)=3\mathrm{x}\mid \mathrm{x}\mid$$

Commented by bemath last updated on 05/Sep/20

$$jooss$$

Answered by 1549442205PVT last updated on 05/Sep/20

$$1)\mathrm{Put}\tan \frac{\mathrm{x}}{2}=\mathrm{t}\Rightarrow \mathrm{dt}=\frac{1}{2}(1+{\mathrm{t}}^2)\mathrm{dx}$$$$\mathrm{cosx}=\frac{1-{\mathrm{t}}^2}{1+{\mathrm{t}}^2},2-\mathrm{cosx}=\frac{{3\mathrm{t}}^2+1}{1+{\mathrm{t}}^2}.\mathrm{Hence},$$$$\mathrm{F}=\int \frac{\cos x}{2-\cos x}dx=2\int \frac{1-{\mathrm{t}}^2}{({3\mathrm{t}}^2+1)(1+{\mathrm{t}}^2)}\mathrm{dt}$$$$=2\int \frac{-({\mathrm{t}}^2+1)+2}{({3\mathrm{t}}^2+1)(1+{\mathrm{t}}^2)}\mathrm{dt}=-2\int \frac{\mathrm{dt}}{{3\mathrm{t}}^2+1}+4\int \frac{\mathrm{dt}}{({3\mathrm{t}}^2+1)({\mathrm{t}}^2+1)}$$$$=\frac{-2}{3}\int \frac{\mathrm{dt}}{{\mathrm{t}}^2+{\left(\frac{1}{\sqrt{3}}\right)}^2}+2\int (\frac{1}{{\mathrm{t}}^2+\frac{1}{3}}-\frac{1}{{\mathrm{t}}^2+1})\mathrm{dt}$$$$=\frac{4}{3}\int \frac{\mathrm{dt}}{{\mathrm{t}}^2+{\left(\frac{1}{\sqrt{3}}\right)}^2}-2\int \frac{\mathrm{dt}}{1+{\mathrm{t}}^2}$$$$=\frac{4}{3}.\frac{\sqrt{3}}{2}{\tan }^{-1}\left(\sqrt{3}\mathrm{t}\right)-{2\tan }^{-1}\left(\mathrm{t}\right)$$$$2)\mathrm{f}\left(\mathrm{x}\right)=\mid {\mathrm{x}}^3\mid \Rightarrow \mathrm{f}{}^2\left(\mathrm{x}\right)={\mathrm{x}}^6.\mathrm{Derivative}\mathrm{two}$$$$\mathrm{sides}\mathrm{by}\mathrm{x}\mathrm{we}\mathrm{get}$$$$2\mathrm{f}\left(\mathrm{x}\right).\mathrm{f}{}^{\prime }\left(\mathrm{x}\right)={6\mathrm{x}}^5\Rightarrow \mathrm{f}{}^{\prime }\left(\mathrm{x}\right)=\frac{{3\mathrm{x}}^5}{\mathrm{f}\left(\mathrm{x}\right)}=\frac{{3\mathrm{x}}^5}{\mid {\mathrm{x}}^3\mid }$$

Answered by Dwaipayan Shikari last updated on 05/Sep/20

$$-\int \frac{-cosx}{2-cosx}=-\int \frac{2-cosx}{2-cosx}-\frac{2}{2-cosx}=-x+2.2\int \frac{1}{2-\frac{1-t^2}{1+t^2}}.\frac{1}{1+t^2}$$$$=-x+4\int \frac{1}{3t^2+1}dt(t=tan\frac{x}{2})$$$$=-x+\frac{4}{3}\int \frac{1}{{\left(t\right)}^2+{\left(\frac{1}{\sqrt{3}}\right)}^2}dt$$$$=-x+\frac{4}{\sqrt{3}}{tan}^{-1}\left(\sqrt{3}tan\frac{x}{2}\right)+C$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com