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Step1/3a. let,Hence, ∫0π​(sinx+cosx)dx = 2Step2/3a. let,Hence, ∫0π​(sinx+cosx)dx = 2b. ∫12​(x+sin4x)dxLet,Now let's consider ∫12​(sin4x)dx = BLet's put 4x = ti.e. 4dx = dti.e. dx =1/4 dtsubstituting we get,But t = 4xCombine    and   .  Combine    and   .  Multiply   .  The result can be shown in multiple forms.Exact Form:  Decimal Form:   Hence, ∫12​(x+sin4x)dx = 1.5018Step3/3a. let,Hence, ∫0π​(sinx+cosx)dx = 2b. ∫12​(x+sin4x)dxLet,Now let's consider ∫12​(sin4x)dx = BLet's put 4x = ti.e. 4dx = dti.e. dx =1/4 dtsubstituting we get,But t = 4xCombine    and   .  Combine    and   .  Multiply   .  The result can be shown in multiple forms.Exact Form:  Decimal Form:   Hence, ∫12​(x+sin4x)dx = 1.5018d. ∫0π/2​(6cos4x+4sin6x)exdx = I i.e. I = I1 + I2Solving I1,and solving I2, we get,Explanation:Hence, the solutions of a,b, and d are given ...