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Category: Integration

1-sinx-cosx-dx-

Question Number 114072 by Lordose last updated on 17/Sep/20 1sinx+cosxdx Answered by Olaf last updated on 17/Sep/20 x=π4usinx+cosx=$$\left(\mathrm{sin}\frac{\pi}{\mathrm{4}}\mathrm{cos}{u}−\mathrm{sin}{u}\mathrm{cos}\frac{\pi}{\mathrm{4}}\right)+\left(\mathrm{cos}\frac{\pi}{\mathrm{4}}\mathrm{cos}{u}+\mathrm{sin}\frac{\pi}{\mathrm{4}}\mathrm{sin}{u}\right) \