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Question Number 201224 by Calculusboy last updated on 02/Dec/23
∫11+sinxdx
Answered by witcher3 last updated on 02/Dec/23
1+sin(x)=hint:∣sin(x2)+cos(x2)∣
Answered by Sutrisno last updated on 04/Dec/23
=∫1sin212x+cos212x+2sin12xcos12xdx=∫1(sin12x+cos12x)2dx=∫1sin12x+cos12xdx=∫12cos(12x−π4)dx=∫sec(12x−π4)2dx=22ln∣sec(12x−π4)+tan(12x−π4)∣+c
Commented by Calculusboy last updated on 04/Dec/23
thankssir
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