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Question-143570




Question Number 143570 by cesarL last updated on 15/Jun/21
Answered by Ar Brandon last updated on 15/Jun/21
I=∫tan^2 8xsec^4 8xdx    =∫tan^2 8xsec^2 8x∙sec^2 8xdx    =(1/8)∫tan^2 8x(1+tan^2 8x)(8sec^2 8x)dx    =(1/8)∫(tan^2 8x+tan^4 8x)d(tan8x)    =((tan^3 8x)/(24))+((tan^5 8x)/(40))+C
$$\mathrm{I}=\int\mathrm{tan}^{\mathrm{2}} \mathrm{8xsec}^{\mathrm{4}} \mathrm{8xdx} \\ $$$$\:\:=\int\mathrm{tan}^{\mathrm{2}} \mathrm{8xsec}^{\mathrm{2}} \mathrm{8x}\centerdot\mathrm{sec}^{\mathrm{2}} \mathrm{8xdx} \\ $$$$\:\:=\frac{\mathrm{1}}{\mathrm{8}}\int\mathrm{tan}^{\mathrm{2}} \mathrm{8x}\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \mathrm{8x}\right)\left(\mathrm{8sec}^{\mathrm{2}} \mathrm{8x}\right)\mathrm{dx} \\ $$$$\:\:=\frac{\mathrm{1}}{\mathrm{8}}\int\left(\mathrm{tan}^{\mathrm{2}} \mathrm{8x}+\mathrm{tan}^{\mathrm{4}} \mathrm{8x}\right)\mathrm{d}\left(\mathrm{tan8x}\right) \\ $$$$\:\:=\frac{\mathrm{tan}^{\mathrm{3}} \mathrm{8x}}{\mathrm{24}}+\frac{\mathrm{tan}^{\mathrm{5}} \mathrm{8x}}{\mathrm{40}}+\mathrm{C} \\ $$

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