Evaluate the integral. 1−sin⁡xcos⁡xdx

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Evaluate the integral.   1−sin⁡xcos⁡xdx

SoosteethicU · Accepted Answer

Given,  1−sin⁡xcos⁡xdx  On simplification, we get  1−sin⁡xcos⁡xdx=∫(1cos⁡x−sin⁡xcos⁡x)  =∫(sec⁡x−tan⁡x)dx  =ln|sec⁡x+tan⁡x|−ln|sec⁡x|+C  =ln|sec⁡x+tan⁡xsec⁡x|+C  =ln|1+tan⁡xsec⁡x|+C  =ln|1+sin⁡xcos⁡x1cos⁡x|+C  =ln|1+sin⁡x|+C

Evaluate the integral. 1−sin⁡xcos⁡xdx

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