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Category: Integration

K-1-1-sin-2-x-dx-

Question Number 149608 by puissant last updated on 06/Aug/21 ..K=11+sin2(x)dx Answered by ArielVyny last updated on 06/Aug/21 sin2x=1cos(2x)2$${K}=\int\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}}}=\int\frac{\mathrm{1}}{\frac{\mathrm{2}+\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}}}{dx} \