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Question Number 9520 by ridwan balatif last updated on 12/Dec/16

$$\int \frac{\mathrm{sinxcosx}}{4+{\sin }^4\mathrm{x}}\mathrm{dx}=...?$$

Answered by mrW last updated on 12/Dec/16

$$\mathrm{u}=\sin \mathrm{x}$$$$\mathrm{du}=\cos \mathrm{xdx}$$$$\int \frac{\mathrm{sinxcosx}}{4+{\sin }^4\mathrm{x}}\mathrm{dx}=\int \frac{\mathrm{u}}{4+{\mathrm{u}}^4}\mathrm{du}$$$$\mathrm{v}={\mathrm{u}}^2$$$$\mathrm{dv}=2\mathrm{udu}$$$$\int \frac{\mathrm{u}}{4+{\mathrm{u}}^4}\mathrm{du}=\frac{1}{2}\int \frac{\mathrm{dv}}{2^2+{\mathrm{v}}^2}=\frac{1}{2}\times \frac{1}{2}\tan {}^{-1}\left(\frac{\mathrm{v}}{2}\right)+\mathrm{C}$$$$=\frac{1}{4}\tan {}^{-1}\left(\frac{{\mathrm{u}}^2}{2}\right)+\mathrm{C}$$$$=\frac{1}{4}\tan {}^{-1}\left(\frac{\sin {}^2\mathrm{x}}{2}\right)+\mathrm{C}$$

Commented by ridwan balatif last updated on 12/Dec/16

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

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