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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} \

x-4-x-4-1-dx-

Question Number 84035 by M±th+et£s last updated on 08/Mar/20 x4x41dx Commented by MJS last updated on 08/Mar/20 $$\frac{{x}−\mathrm{4}}{{x}^{\mathrm{4}} −\mathrm{1}}=−\frac{{x}}{\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{1}\right)}+\frac{\mathrm{2}}{{x}^{\mathrm{2}} +\mathrm{1}}+\frac{\mathrm{5}}{\mathrm{4}\left({x}+\mathrm{1}\right)}−\frac{\mathrm{3}}{\mathrm{4}\left({x}−\mathrm{1}\right)} \