Question

First make a substitution and then use IBP to evaluate the integral -4t^3*cos(t^2)dt from the sqr(pi/2) to the sqr (pi)?

Answer 1

I let u=t^2 and du=2tdt and dt=du/2t.

Answer 2

use t^2 as z

Answer 3

yup going in correct way

Answer 4

integral of -4t^3*cos(t^2)dt if u=t^2du=2tdt
then question can be written as -(2t^2)(2t) cos(t^2)dt
the question transforms to -2ucosudu and the n use integration by parts

Answer 5

Okay, thank you!