Resolução de Integral Trigonométrica

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18:43 Given: π/2 sin (x I x (tan dt exp In In du - 6 dx Solution: π/2 dt Step Evaluate Step 2: Evaluate In In (1-u) du 0 1+ (tan Let = tan t du = sec2 = = = dt Let I1 = In In du dt du Using the known result: When t=0 = when t = π/2 dt 8 1 du du = Inx In dx = da = 1+ (tant) Let => u = du = 1/12 -1 dv, = For a 1, , this gives I1 = 6 Therefore, 1 8 1 = dv 1 = dv 2+ π² V2 (1+v) II This is a standard Beta-type integral: Step 3: Hence, π dx = q sin 6 exp = 1 Here P = , q = Step 4: Now the given integral becomes π π dv = = sin I = sin dx π/2 Using the standard result: Hence, dt 1 π = = 1+ V2 2 sin x (ax) dx = 2 π/2 dt π Here a = 2 >0 I = π 2 = 1+ (tan 2 I = 2 + Adicione sua própria foto zabih khan 26 Respondendo a zabih khan