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Q.

∫cosec2x-2018cos2018x dx=acosbx·sincx+k then abc=

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answer is -2017.

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Detailed Solution

Given ∫cosec2x-2018cos2018x d=∫cosec2x·sec2018x dx-∫2018·sec2018x dx                                                 =sec2018x·(-cotx)+∫2018·sec2017x·secx·tanx·cotx dx-∫2018·sec2018x dx   integrating by parts                                                  =-1cos2018x·cosxsinx+k                                                  =-1cos2017x·sinx+k                                                  =acosbx·sincx+k   (given) ∴a=-1,b=2017,c=1∴abc=(-1)(2017)(1)=-2017
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