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Question Number 186171 by normans last updated on 01/Feb/23

                 [so easy]         ∫ cos^2  (4x) + sin^4  (2x) dx

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[\boldsymbol{{so}}\:\boldsymbol{{easy}}\right] \\ $$$$\:\:\:\:\:\:\:\int\:\boldsymbol{{cos}}^{\mathrm{2}} \:\left(\mathrm{4}\boldsymbol{{x}}\right)\:+\:\boldsymbol{{sin}}^{\mathrm{4}} \:\left(\mathrm{2}\boldsymbol{{x}}\right)\:\boldsymbol{{dx}}\:\:\: \\ $$$$ \\ $$

Answered by MJS_new last updated on 02/Feb/23

easy−cheesy:  ∫cos^2  4x +sin^4  2x dx=  =(5/(64))sin 8x −(1/8)sin 4x +(7/8)x+C

$$\mathrm{easy}−\mathrm{cheesy}: \\ $$$$\int\mathrm{cos}^{\mathrm{2}} \:\mathrm{4}{x}\:+\mathrm{sin}^{\mathrm{4}} \:\mathrm{2}{x}\:{dx}= \\ $$$$=\frac{\mathrm{5}}{\mathrm{64}}\mathrm{sin}\:\mathrm{8}{x}\:−\frac{\mathrm{1}}{\mathrm{8}}\mathrm{sin}\:\mathrm{4}{x}\:+\frac{\mathrm{7}}{\mathrm{8}}{x}+{C} \\ $$

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