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Solve integral of    [(cos 2 2x) / cos 2 x ]  dx

7 years ago


Answers : (1)


{cos2x/cos(x)}^2 = {(2cos^2(x)-1)/cosx}^2

=> {2 cosx-secx}^2 = {4(cosx)^2+(secx)^2-4}

=> {2{2(cosx)^2 -1} -2 +(secx^2)}

=> 2cos(2x)dx -2dx +9secx)^2dx

=> integral of {cos2x/cos(x)}^2 = integral of 2cos(2x)dx -2dx +9secx)^2dx

= sin(2x) -2x +tan(x)+c.
7 years ago

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