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Question Number 140998 by mohammad17 last updated on 14/May/21

Answered by Ar Brandon last updated on 14/May/21

$$\mathcal{I}=\int \frac{\sin 4\mathrm{x}}{{\cos }^4\mathrm{x}+4}\mathrm{dx}=\int \frac{2\sin 2\mathrm{xcos}2\mathrm{x}}{{\left(\frac{1+\cos 2\mathrm{x}}{2}\right)}^2+4}\mathrm{dx},\mathrm{c}=\cos 2\mathrm{x}$$$$=-\int \frac{\mathrm{c}}{\frac{{(\mathrm{c}+1)}^2}{4}+4}\mathrm{dc}=-4\int \frac{\mathrm{cdc}}{{\mathrm{c}}^2+2\mathrm{c}+17}\mathrm{dc}$$

Commented by mohammad17 last updated on 14/May/21

$$chose\left(a\right)or\left(b\right)or\left(c\right)or\left(d\right)$$$$$$$$ithinkd$$

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