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Question Number 26952 by Femmy last updated on 31/Dec/17
Answered by prakash jain last updated on 31/Dec/17
∫sin5xcos3xdx=∫sin4xcos3xsinxdx=∫(1−cos2x)2cos3xsinxdx=∫1−2cos2x+cos4xcos3xsinxdxcosx=u⇒−sinxdx=du=−∫(1u3−2u+u)du=−[−12u2−2lnu+u22]+c=−[−12cos2x−2lncosx+cos2x2]+c
Commented by Femmy last updated on 31/Dec/17
thankssirbutcheckyouransagainlittlemistake8o
Commented by prakash jain last updated on 31/Dec/17
Thanks.corrected
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