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Question

Evaluate sec2x.cosec2xdx on IR ({nπ;nZ}{(2n+1)π2:nZ})

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Solution

sec2xcosec2xdx
=sec2x(1+cot2x)dx
=sec2x+sec2x.cot2xdx
=sec2x.dx+1cos2x.cos2xsin2xdx
=tanx+cosec2xdx
=tanxcotx+c

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