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 114094 by Lordose last updated on 17/Sep/20

$$\int \mathbf{ln}\left(\mathbf{x}\right){\mathbf{sin}}^{-1}\left(\mathbf{x}\right)\mathbf{dx}$$

Answered by Olaf last updated on 17/Sep/20

$$u^{\prime }=\ln x,u=x\ln x-x$$$$v=\arcsin x,v^{\prime }=\frac{1}{\sqrt{1-x^2}}$$$$(x\ln x-x)\arcsin x-\int \frac{x\ln x-x}{\sqrt{1-x^2}}dx$$$$x\ln \left(\frac{x}{e}\right)\arcsin x+\int \ln \left(\frac{x}{e}\right)\frac{-2x}{2\sqrt{1-x^2}}dx$$$$u=\ln \left(\frac{x}{e}\right),u^{\prime }=\frac{1}{x}$$$$v^{\prime }=\frac{-2x}{2\sqrt{1-x^2}},v=\sqrt{1-x^2}$$$$x\ln \left(\frac{x}{e}\right)\arcsin x+\ln \left(\frac{x}{e}\right)\sqrt{1-x^2}-\int \frac{1}{x}\sqrt{1-x^2}dx$$$$\ln \left(\frac{x}{e}\right)(x\arcsin x+\sqrt{1-x^2})-\int \frac{1}{x}\sqrt{1-x^2}dx$$$$x=\mathrm{sinu}$$$$\int \frac{1}{x}\sqrt{1-x^2}dx=$$$$\int \frac{1}{\sin u}\sqrt{1-{\sin }^{2}u}.\cos udu=$$$$\int \epsilon \frac{{\cos }^{2}u}{\sin u}du=\epsilon \int \frac{1-{\sin }^{2}u}{\sin u}du(\epsilon =\pm 1)$$$$=\epsilon \int (\csc u-\sin u)du=$$$$\epsilon (\frac{1}{2}\ln \mid \frac{1-\cos u}{1+\cos u}\mid +\cos u)=$$$$\epsilon (\frac{1}{2}\ln \mid \frac{1-\sqrt{1-x^2}}{1+\sqrt{1-x^2}}\mid +\sqrt{1-x^2})...$$$$$$

Commented by Lordose last updated on 17/Sep/20

$$\mathbf{Complete}\mathbf{it}\mathbf{sir}$$

Answered by 1549442205PVT last updated on 17/Sep/20

$$\mathrm{Integrating}\mathrm{by}\mathrm{part}\mathrm{we}\mathrm{have}$$$$\mathrm{I}=\int \mathbf{ln}\left(\mathbf{x}\right){\mathbf{sin}}^{-1}\left(\mathbf{x}\right)\mathbf{dx}=\int (\mathrm{lnx}-1){\sin }^{-1}\left(\mathrm{x}\right)\mathrm{dx}+\int {\sin }^{-1}\left(\mathrm{x}\right)\mathrm{dx}$$$$=(\mathrm{xlnx}-\mathrm{x}){\sin }^{-1}\left(\mathrm{x}\right)-\int \mathrm{x}{\left[(\mathrm{lnx}-1){\sin }^{-1}\left(\mathrm{x}\right)\right]}^{\prime }\mathrm{dx}+\int {\sin }^{-1}\mathrm{xdx}$$$$=(\mathrm{xlnx}-\mathrm{x})-\int \frac{\mathrm{xlnx}}{\sqrt{1-{\mathrm{x}}^2}}\mathrm{dx}+\int \frac{\mathrm{x}}{\sqrt{1-{\mathrm{x}}^2}}\mathrm{dx}\left(1\right)$$$$\mathrm{Since}\int \frac{\mathrm{x}}{\sqrt{1-{\mathrm{x}}^2}}\mathrm{dx}=\sqrt{1-{\mathrm{x}}^2},$$$$\mathrm{We}\mathrm{just}\mathrm{need}\mathrm{must}\mathrm{find}\int \frac{\mathrm{xlnx}}{\sqrt{1-{\mathrm{x}}^2}}\mathrm{dx}$$$$\mathrm{Put}\mathrm{x}=\sin \theta \Rightarrow \mathrm{dx}=\cos \theta \mathrm{d}\theta$$$$\int \frac{\mathrm{xlnx}}{\sqrt{1-{\mathrm{x}}^2}}\mathrm{dx}=\int \sin \theta \mathrm{lnsin}\theta \mathrm{d}\theta =-\cos \theta \mathrm{lnsin}\theta$$$$-\int (-\cos \theta ){\left(\mathrm{lnsin}\theta \right)}^{\prime }\mathrm{d}\theta =-\cos \theta \mathrm{lnsin}\theta +\int \frac{{\cos }^2\theta }{\sin \theta }\mathrm{d}\theta$$$$=-\cos \theta \mathrm{lnsin}\theta +\int \frac{1-{\sin }^2\theta }{\sin \theta }\mathrm{d}\theta =$$$$=-\cos \theta \mathrm{lnsin}\theta +\int (\mathrm{cosec}\theta -\sin \theta )\mathrm{d}\theta$$$$=-\cos \theta \mathrm{lnsin}\theta +\ln (\mathrm{cosec}\theta -\cos \theta )+\cos \theta$$$$=-\sqrt{1-{\mathrm{x}}^2}\mathrm{lnx}+\ln \left(\frac{1-\sqrt{1-{\mathrm{x}}^2}}{\mathrm{x}}\right)+\sqrt{1-{\mathrm{x}}^2}$$$$\mathrm{From}\left(1\right)\mathrm{we}\mathrm{get}$$$$\mathbf{I}=(\mathbf{x}\mathrm{l}\mathbf{nx}-\mathbf{x}){\mathbf{sin}}^{-1}\mathbf{x}+\sqrt{1-{\mathbf{x}}^2}\mathbf{lnx}-\mathbf{ln}\left(\frac{1-\sqrt{1-{\mathbf{x}}^2}}{\mathbf{x}}\right)+\mathbf{C}$$

Answered by MJS_new last updated on 17/Sep/20

$$\mathrm{again}\mathrm{just}\mathrm{because}\mathrm{something}\mathrm{went}\mathrm{wrong}\mathrm{above}...$$$$\int \ln x\arcsin xdx=$$$$\left[\mathrm{by}\mathrm{parts}\right]$$$$=(\sqrt{1-x^2}+x\arcsin x)\ln x-\int \frac{\sqrt{1-x^2}}{x}+\arcsin xdx$$$$\int \frac{\sqrt{1-x^2}}{x}dx=\sqrt{1-x^2}+\ln \frac{x}{1+\sqrt{1-x^2}}$$$$\int \arcsin xdx=\sqrt{1-x^2}+x\arcsin x$$$$\Rightarrow$$$$\int \ln x\arcsin xdx=$$$$=\ln \frac{1+\sqrt{1-x^2}}{x}+(\sqrt{1-x^2}+x\arcsin x)\ln x-x\arcsin x-2\sqrt{1-x^2}+C$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com