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

simplifying the other Iwe also getNow adding both the I’s we getNow this integral is easily solved

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Integral Calculus> Find I = ∫(3x 2 -3x+1)cos(x 3 -3x 2 +4x-2...

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

Jayant Kishore , 9 Years ago

Grade 12

1 Answers

Riddhish Bhalodia

$I = \int \limits_0^2(3x^2-3x+1)cos(x^3-3x^2+4x-2)dx = \int \limits_0^2(3(2-x)^2-3(2-x)+1)cos((2-x)^3-3(2-x)^2+4(2-x)-2)dx$

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