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Question Number 170722 by Sotoberry last updated on 29/May/22

Answered by MikeH last updated on 29/May/22

$$u=\sin \theta$$$$du=\cos \theta d\theta$$$$dv=e^{\theta }d\theta$$$$v=e^{\theta }$$$$\Rightarrow \int e^{\theta }\sin \theta d\theta =\left[e^{\theta }\sin \theta \right]-\int e^{\theta }\cos \theta d\theta$$$$\int e^{\theta }\cos \theta d\theta =[-e^{\theta }\cos \theta ]+\int e^{\theta }\sin \theta d\theta$$$$\Rightarrow 2\int e^{\theta }\sin \theta d\theta =e^{\theta }\sin \theta -e^{\theta }\cos \theta$$$$\Rightarrow \int e^{\theta }\sin \theta d\theta =\frac{e^{\theta }}{2}(\sin \theta -\cos \theta )$$

Answered by floor(10²Eta[1]) last updated on 29/May/22

$$\mathrm{u}=\sin \theta$$$$\mathrm{du}=\cos \theta \mathrm{d}\theta$$$$\mathrm{dv}={\mathrm{e}}^{\theta }\mathrm{d}\theta$$$$\mathrm{v}={\mathrm{e}}^{\theta }$$$$\mathrm{I}=\int {\mathrm{e}}^{\theta }\sin \theta \mathrm{d}\theta ={\mathrm{e}}^{\theta }\sin \theta -\int {\mathrm{e}}^{\theta }\cos \theta \mathrm{d}\theta$$$$\mathrm{u}=\cos \theta$$$$\mathrm{du}=-\sin \theta \mathrm{d}\theta$$$$\mathrm{dv}={\mathrm{e}}^{\theta }\mathrm{d}\theta$$$$\mathrm{v}={\mathrm{e}}^{\theta }$$$$\mathrm{I}={\mathrm{e}}^{\theta }\sin \theta -({\mathrm{e}}^{\theta }\cos \theta +\int {\mathrm{e}}^{\theta }\sin \theta \mathrm{d}\theta )$$$$\Rightarrow \mathrm{I}={\mathrm{e}}^{\theta }(\sin \theta -\cos \theta )-\mathrm{I}$$$$\mathrm{I}=\frac{{\mathrm{e}}^{\theta }}{2}(\sin \theta -\cos \theta )$$

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