If ∫ex(sin22x−8cos4x)dx=exf(x)+CThen fπ4

∫exsin2⁡2x−8cos⁡4xdx=∫ex1−cos⁡4x2+2sin⁡4x−2sin⁡4x−8cos⁡4xdx=∫ex1−cos⁡4x2+2sin⁡4xdx−2∫ex(4sin⁡4x+4cos⁡4x)dx=ex1−cos⁡4x2−2exsin⁡4x+C=ex1−cos⁡4x2−2sin⁡4x+C⇒f(x)=1−cos⁡4x2−2sin⁡4x∴fπ4=1

If ∫ex(sin22x−8cos4x)dx=exf(x)+CThen fπ4

∫exsin2⁡2x−8cos⁡4xdx=∫ex1−cos⁡4x2+2sin⁡4x−2sin⁡4x−8cos⁡4xdx=∫ex1−cos⁡4x2+2sin⁡4xdx−2∫ex(4sin⁡4x+4cos⁡4x)dx=ex1−cos⁡4x2−2exsin⁡4x+C=ex1−cos⁡4x2−2sin⁡4x+C⇒f(x)=1−cos⁡4x2−2sin⁡4x∴fπ4=1

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If $\int e^{x} (\sin^{2} 2 x - 8 \cos 4 x) d x = e^{x} f (x) + C$
Then $f (\frac{π}{4})$

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Detailed Solution

$\begin{array}{l} \int e^{x} (\sin^{2} 2 x - 8 \cos 4 x) d x \\ = \int e^{x} (\frac{1 - \cos 4 x}{2} + 2 \sin 4 x - 2 \sin 4 x - 8 \cos 4 x) d x \\ = \int e^{x} (\frac{1 - \cos 4 x}{2} + 2 \sin 4 x) d x - 2 \int e^{x} (4 \sin 4 x + 4 \cos 4 x) d x \\ = e^{x} (\frac{1 - \cos 4 x}{2}) - 2 e^{x} \sin 4 x + C \\ = e^{x} (\frac{1 - \cos 4 x}{2} - 2 \sin 4 x) + C \\ \Rightarrow f (x) = \frac{1 - \cos 4 x}{2} - 2 \sin 4 x \\ ∴ f (\frac{π}{4}) = 1 \end{array}$