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Question Number 141387 by cesarL last updated on 18/May/21
∫−π/4π/4(sec2x+tgx)2dx
Answered by MJS_new last updated on 18/May/21
∫π/4−π/4(sec2x+tanx)2dx=[t=tanx→dx=cos2xdt]=∫1−1(t2+t+1)2t2+1dt=∫1−1(t2+2t+2−1t2+1)dt==[t33+t2+2t−arctant]−11=143−π2
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