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Question Number 129564 by bramlexs22 last updated on 16/Jan/21

$$\mathcal{V}=\int \frac{\sin \mathrm{x}}{\sin (\mathrm{x}+\theta )}\mathrm{dx}$$

Answered by Lordose last updated on 16/Jan/21

$$\mathrm{\Omega }=\int \frac{\sin \left(\mathrm{x}\right)}{\sin (\mathrm{x}+\theta )}\mathrm{dx}\stackrel{\mathrm{u}=\mathrm{x}+\theta }{=}\int \frac{\sin (\mathrm{u}-\theta )}{\sin \left(\mathrm{u}\right)}\mathrm{du}$$$$\mathrm{\Omega }=\int \frac{\sin \left(\mathrm{u}\right)\cos \theta -\cos \left(\mathrm{u}\right)\sin \theta }{\sin \left(\mathrm{u}\right)}\mathrm{du}=\mathrm{ucos}\theta -\sin \theta \ln \left(\mathrm{sinu}\right)+\mathrm{C}$$$$\mathrm{\Omega }=(\mathrm{x}+\theta )\cos \theta -\sin \theta \ln \left(\sin (\mathrm{x}+\theta )\right)+\mathrm{C}$$

Answered by liberty last updated on 16/Jan/21

$$\mathcal{V}=\int \frac{\sin ((\mathrm{x}+\theta )-\theta )}{\sin (\mathrm{x}+\theta )}\mathrm{dx}$$$$=\int \frac{\sin (\mathrm{x}+\theta )\cos \theta -\cos (\mathrm{x}+\theta )\sin \theta }{\sin (\mathrm{x}+\theta )}\mathrm{dx}$$$$=\int \cos \theta \mathrm{dx}-\sin \theta \int \frac{\cos (\mathrm{x}+\theta )}{\sin (\mathrm{x}+\theta )}\mathrm{dx}$$$$=\mathrm{xcos}\theta -\sin \theta .\ln \mid \sin (\mathrm{x}+\theta )\mid +\mathrm{C}$$

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