Question

inegral of root tanx - root cotx

Answer 1

∫π20tanx−−−−√dx+cotx−−−−√dx∫0π2tan⁡xdx+cot⁡xdx =∫π20sinx+cosxsinxcosx−−−−−−−−√dx=∫π20sinx+cosx2sinxcosx√2√dx=2–√∫π20sinx+cosx1−(1−2sinxcosx)−−−−−−−−−−−−−−−−−√dx=∫0π2sin⁡x+cos⁡xsin⁡xcos⁡xdx=∫0π2sin⁡x+cos⁡x2sin⁡xcos⁡x2dx=2∫0π2sin⁡x+cos⁡x1−(1−2sin⁡xcos⁡x)dx =2–√∫π20sinx+cosx1−(sinx−cosx)2−−−−−−−−−−−−−−−√dx=2∫0π2sin⁡x+cos⁡x1−(sin⁡x−cos⁡x)2dx

Let t=sinx−cosxt=sin⁡x−cos⁡x , dx=dtsinx+cosxdx=dtsin⁡x+cos⁡x

x→π2⟹t=(sinx−cosx)→1x→π2⟹t=(sin⁡x−cos⁡x)→1 x→0⟹t=(sinx−cosx)→−1x→0⟹t=(sin⁡x−cos⁡x)→−1

2–√∫1−111−t2−−−−−√dt=2–√[sin−1t]1−1=2–√[π2−(−π2)]=2–√π2∫−1111−t2dt=2[sin−1⁡t]−11=2[π2−(−π2)]=2π
hope it is use full