Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

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 

Grade:12

1 Answers

Riddhish Bhalodia
askIITians Faculty 434 Points
5 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

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free