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Question Number 83961 by mathmax by abdo last updated on 08/Mar/20

$$findI=\int e^{-x}{cos}^{4}xdxandJ=\int e^{-x}{sin}^{4}xdx$$

Commented by mathmax by abdo last updated on 11/Mar/20

$$wehaveI+J=\int e^{-x}({cos}^{4}x+{sin}^{4}x)dx$$$$=\int e^{-x}({({cos}^{2}x+{sin}^{2}x)}^2-2{cos}^{2}x{sin}^{2}x)dx$$$$=\int e^{-x}(1-2{\left(\frac{sin\left(2x\right)}{2}\right)}^2)dx=\int e^{-x}(1-\frac{1}{2}{sin}^2\left(2x\right))dx$$$$=\int e^{-x}dx-\frac{1}{2}\int e^{-x}\frac{1-cos\left(4x\right)}{2}dx$$$$=-e^{-x}-\frac{1}{4}\int e^{-x}dx+\frac{1}{4}\int e^{-x}cos\left(4x\right)dx$$$$=(-1+\frac{1}{4})e^{-x}+\frac{1}{4}Re(\int e^{-x+i4x}dx)$$$$\int e^{(-1+4i)x}dx=\frac{1}{-1+4i}{e}^{(-1+4i)x}=-\frac{1}{1-4i}{e}^{(-1+4i)x}$$$$=-\frac{1+4i}{17}{e}^{-x}(cos\left(4x\right)+isin\left(4x\right))$$$$=-\frac{e^{-x}}{17}(cos\left(4x\right)+isin\left(4x\right)+4icos\left(4x\right)-4sin\left(4x\right))\Rightarrow$$$$I+J=-\frac{3}{4}{e}^{-x}-\frac{e^{-x}}{68}(cos\left(4x\right)-4sin\left(4x\right))$$$$I-J=\int e^{-x}({cos}^{4}x-{sin}^{2}x)dx=\int e^{-x}({cos}^{2}x-{sin}^{2}x)dx$$$$=\int e^{-x}cos\left(2x\right)dx=Re(\int e^{(-1+2i)x}dx)$$$$wehave\int e^{(-1+2i)x}dx=\frac{1}{-1+2i}{e}^{(-1+2i)x}$$$$=-\frac{1+2i}{5}{e}^{-x}(cos\left(2x\right)+isin\left(2x\right))$$$$=-\frac{e^{-x}}{5}(cos\left(2x\right)+isin\left(2x\right)+2icos\left(2x\right)-2sin\left(2x\right))\Rightarrow$$$$I-J=-\frac{e^{-x}}{5}(cos\left(2x\right)-2sin\left(2x\right))$$$$nowitseazytodetermineIandJ$$

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