Find u and du The indefinite integral sec2xtan2xdx I need help telling the derivative from the function apart.
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Originally Posted by saraduh Find u and du The indefinite integral sec2xtan2xdx I need help telling the derivative from the function apart. Try $\displaystyle u = \sec{2x}$
Would this work? sec2xtan2xdx = sin2x dx/(cos2x)^2 let u = cos2x---> du = -2sin2xdx
Originally Posted by saraduh Would this work? Mr F says: Yes. Good idea. sec2xtan2xdx = sin2x dx/(cos2x)^2 let u = cos2x---> du = -2sin2xdx ..
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