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B
esinx(x−secx)+C
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C
esinx(secx+tanx)+C
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D
Noneofthese
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Solution
The correct option is Besinx(x−secx)+C We have ∫esinx(xcos3x−sinxcos2x)dx=∫xesinxcosxdx−∫esinx(secxtanx)dx=[xesinx−∫esinxdx]−[esinxsecx−∫esinxdx]=esinx(x−secx)+C