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 84766 by jagoll last updated on 15/Mar/20

$$\int \frac{\mathrm{dx}}{{(16+9\sin \mathrm{x})}^2}$$$$$$

Commented by jagoll last updated on 16/Mar/20

$$\int \sec \mathrm{x}\left[\frac{\cos \mathrm{x}}{{(16+9\sin \mathrm{x})}^2}\right]\mathrm{dx}=$$$$$$$$\mathrm{u}=\sec \mathrm{x}\Rightarrow \mathrm{du}=\sec \mathrm{x}\tan \mathrm{x}\mathrm{dx}$$$$\mathrm{v}=\frac{1}{9}\int \frac{\mathrm{d}(16+9\sin \mathrm{x})}{{(16+9\sin \mathrm{x})}^2}=-\frac{1}{16+9\sin \mathrm{x}}$$$$\Rightarrow -\frac{\sec \mathrm{x}}{16+9\sin \mathrm{x}}+\int \frac{\sec \mathrm{x}\tan \mathrm{x}}{16+9\sin \mathrm{x}}\mathrm{dx}$$$${\mathrm{I}}_2=\int \frac{\sin \mathrm{x}\mathrm{dx}}{\cos {}^2\mathrm{x}(16+9\sin \mathrm{x})}$$

Commented by abdomathmax last updated on 16/Mar/20

$$letf\left(a\right)=\int \frac{dx}{a+9sinx}wehave{f}^{{}^{\prime }}\left(a\right)=-\int \frac{dx}{{(a+9sinx)}^2}$$$$\Rightarrow \int \frac{dx}{{(a+9sinx)}^2}=-f^{{}^{\prime }}\left(a\right)$$$$f\left(a\right){=}_{tan\left(\frac{x}{2}\right)=t}\int \frac{2dt}{(1+t^2)(a+9\times \frac{2t}{1+t^2})}$$$$=\int \frac{2dt}{a+{at}^2+18t}=\int \frac{2dt}{{at}^2+18t+a}$$$${\mathrm{\Delta }}^{{}^{\prime }}=9^2-a^{2}wetakea>9\Rightarrow t_1=\frac{-9+\sqrt{81-a^2}}{a}$$$$t_2=\frac{-9-\sqrt{81-a^2}}{a}\Rightarrow f\left(a\right)=\int \frac{2dt}{a(t-t_1)(t-t_2)}$$$$=\frac{2}{a\times \frac{2\sqrt{81-a^2}}{a}}\int (\frac{1}{t-t_1}-\frac{1}{t-t_2})dt$$$$=\frac{1}{\sqrt{81-a^2}}ln\mid \frac{t-t_1}{t-t_2}\mid +C$$$$=\frac{1}{\sqrt{81-a^2}}ln\mid \frac{t-\frac{-9+\sqrt{81-a^2}}{a}}{t-\frac{-9-\sqrt{81-a^2}}{a}}\mid +C$$$$=\frac{1}{\sqrt{81-a^2}}ln\mid \frac{at+9-\sqrt{81-a^2}}{at+9+\sqrt{81-a^2}}\mid +C$$$$restcalculusof{f}^{{}^{\prime }}\left(a\right)...becontinued...$$

Answered by TANMAY PANACEA last updated on 16/Mar/20

$$t=a+bsinx\to sinx=\frac{t-a}{b}$$$$p=\frac{cosx}{a+bsinx}$$$$\frac{dp}{dx}=\frac{(a+bsinx)\times -sinx-cosx\left(bcosx\right)}{{(a+bsinx)}^2}$$$$\frac{dp}{dx}=\frac{-asinx-{bsin}^{2}x-{bcos}^{2}x}{{(a+bsinx)}^2}=\frac{-(asinx+b)}{{(a+bsinx)}^2}$$$$nowputtingsinx=\frac{t-a}{b}$$$$-\frac{dp}{dx}=\frac{(a\times \frac{t-a}{b}+b)}{t^2}=\frac{at-a^2+b^2}{{bt}^2}=\frac{a}{b}\times \frac{1}{t}+\frac{b^2-a^2}{t^2}$$$$-\frac{\mathit{d}\mathit{p}}{\mathit{d}\mathit{x}}=\frac{\mathit{a}}{\mathit{b}}\times \frac{1}{\mathit{t}}+\frac{{\mathit{b}}^2-{\mathit{a}}^2}{{\mathit{t}}^2}$$$$\frac{-1}{({\mathit{b}}^2-{\mathit{a}}^2)}\frac{\mathit{d}\mathit{p}}{\mathit{d}\mathit{x}}=\frac{\mathit{a}}{\mathit{b}({\mathit{b}}^2-{\mathit{a}}^2)}\times \frac{1}{(\mathit{a}+\mathit{b}\mathit{s}\mathit{i}\mathit{n}\mathit{x})}+\frac{1}{{(\mathit{a}+\mathit{b}\mathit{s}\mathit{i}\mathit{n}\mathit{x})}^2}$$$$\frac{1}{{\mathit{a}}^2-{\mathit{b}}^2}\times \mathit{d}\left(\frac{\mathit{c}\mathit{o}\mathit{s}\mathit{x}}{\mathit{a}+\mathit{b}sinx}\right)=\frac{\mathit{a}}{\mathit{b}({\mathit{b}}^2-{\mathit{a}}^2)}\times \frac{\mathit{d}\mathit{x}}{(\mathit{a}+\mathit{b}\mathit{s}\mathit{i}\mathit{n}\mathit{x})}+\frac{\mathit{d}\mathit{x}}{{(\mathit{a}+\mathit{b}\mathit{s}\mathit{i}\mathit{n}\mathit{x})}^2}$$$$\mathit{n}\mathit{o}\mathit{w}\int \frac{dx}{{(a+bsinx)}^2}$$$$=\frac{1}{a^2-b^2}\int d\left(\frac{cosx}{a+bsinx}\right)-\frac{a}{b(b^2-a^2)}\int \frac{dx}{a+bsinx}$$$$a=16b=9$$$$\int \frac{dx}{{(16+9sinx)}^2}=\frac{1}{({16}^2-9^2)}\int d\left(\frac{cosx}{16+9sinx}\right)+\frac{16}{9({16}^2-9^2)}\int \frac{dx}{16+9sinx}$$$$now$$$$\int \frac{dx}{16+9sinx}k=tan\frac{x}{2}\to dk={sec}^2\frac{x}{2}\times \frac{1}{2}dx\to \frac{2dk}{1+k^2}=dx$$$$\int \frac{2dk}{(1+k^2)}\times \frac{1}{(16+9\times \frac{2k}{1+k^2})}$$$$2\int \frac{dk}{16+16k^2+18k}=\frac{1}{8}\int \frac{dk}{k^2+\frac{9k}{8}+1}=\frac{1}{8}\int \frac{dk}{k^2+2.k.\frac{9}{16}+{\left(\frac{9}{16}\right)}^2+1-{\left(\frac{9}{16}\right)}^2}$$$$=\frac{1}{8}\int \frac{dk}{{(k+\frac{9}{16})}^2+\frac{175}{{16}^2}}=\frac{1}{8}\times \int \frac{dk}{{(k+\frac{9}{16})}^2+{\left(\frac{5\sqrt{7}}{16}\right)}^2}$$$$=\frac{1}{8}\times \frac{16}{5\sqrt{7}}{tan}^{-1}\left(\frac{k+\frac{9}{16}}{\frac{5\sqrt{7}}{16}}\right)=\frac{2}{5\sqrt{7}}{tan}^{-1}\left(\frac{16k+9}{5\sqrt{7}}\right)$$$$\mathit{s}\mathit{o}\mathit{a}\mathit{n}\mathit{s}\mathit{w}\mathit{e}\mathit{r}\mathit{i}\mathit{s}$$$$\frac{1}{({16}^2-9^2)}\times \left(\frac{cosx}{16+9sinx}\right)+\frac{16}{9({16}^2-9^2)}\times \frac{2}{5\sqrt{7}}{tan}^{-1}\left(\frac{16tan\frac{x}{2}+9}{5\sqrt{7}}\right)+c$$$$$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com