How do you integrate #f(x)=intsin(e^t)dt# between 4 to #x^2#?

1 Answer
May 22, 2018

Answer:

#int_4^(x^2)sin(e^t)dt≈int _0^(+oo)sin(y)/ydy=pi/2#

Explanation:

#int_4^(x^2)sin(e^t)dt#
Let #y=e^t#
#t=lny#
#dt=(dy)/y#
#int_4^(x^2)sin(e^t)dt=int_(e^4)^(e^(x²))sin(y)/ydy#, and we can see that there's no result of this integration. However, using Dirichlet integral:
#(int_(e^4)^(e^(x²))sin(y)/ydy) _(x to oo)≈int _0^(+oo)sin(y)/ydy=pi/2#
\0/ Here's our answer!