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Question Number 60346 by necx1 last updated on 20/May/19
∫xsec3xdxpleasehelp
Commented by kaivan.ahmadi last updated on 20/May/19
first∫sec3xdx=(tgxcosx+ln∣secx+tgx∣)/2nowletu=x⇒du=dxdv=sec3xdx⇒v=(tgxcosx+ln∣secx+tgx∣)/2uv−∫vdu=x(tgxcosx+lb∣secx+tgx∣)/2−12∫(tgxcosx+ln∣secx+tgx∣)dxontheotherhand∫tgxcosxdx=∫sinxcos2xdx=1cosx=secxand∫ln∣secx+tgx∣dx=xln(secx+tgx)−secx−tgx
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