Find I = ∫(3x 2 -3x+1)cos(x 3 -3x 2 +4x-2)dx form lower limit 0 to upper limit 2

Find I = ∫(3x2-3x+1)cos(x3-3x2+4x-2)dx  form lower limit 0 to upper limit 2 


1 Answers

Riddhish Bhalodia
askIITians Faculty 434 Points
6 years ago
I = \int \limits_0^2(3x^2-3x+1)cos(x^3-3x^2+4x-2)dx = \int \limits_0^2(3(2-x)^2-3(2-x)+1)cos((2-x)^3-3(2-x)^2+4(2-x)-2)dx

simplifying the other I
we also get
I = \int \limits_0^2(3x^2-9x+7)cos(-(x^3-3x^2+4x-2))dx
Now adding both the I’s we get
2I = 2\int \limits_0^2(3x^2-6x+4)cos(x^3-3x^2+4x-2)dx
Now this integral is easily solved

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