Question Number 80293 by ~blr237~ last updated on 01/Feb/20

Commented by john santu last updated on 02/Feb/20

Commented by ~blr237~ last updated on 02/Feb/20

Commented by john santu last updated on 02/Feb/20
![(d/dx)[∫_0 ^x f(u)cos (x−u)du]= f(x)cos (x)+f(0) cos (0) ?](https://www.tinkutara.com/question/Q80308.png)
Commented by ~blr237~ last updated on 02/Feb/20

Commented by jagoll last updated on 02/Feb/20

Commented by ~blr237~ last updated on 02/Feb/20

Commented by jagoll last updated on 02/Feb/20

Commented by ~blr237~ last updated on 02/Feb/20
![sir what is going on? let take f(t)=t F(x)=∫_0 ^x (x−t)cos2tdt by part u′=cos2t and v=x−t F(x)=[(1/2)(x−t)sin2t]_0 ^x +(1/2)∫_0 ^x sin2tdt =(1/2)[((−cos2t)/2)]_0 ^x =((1−cos2x)/4) (dF/dx)=((sin2x)/2) ≠ f(x)=x](https://www.tinkutara.com/question/Q80323.png)
Commented by jagoll last updated on 02/Feb/20

Commented by ~blr237~ last updated on 02/Feb/20
![When you want to derivate F(x)=∫_0 ^(u(x)) f(x,t)dt make sure firstly (by a good stating variable: may be z=(t/(u(x))) ) to remove x at the whole boundary we were having F(x)=∫_0 ^x f(x−t)cos2tdt F(x)=∫_0 ^1 x[f(x−xv)cos(2xv)]dv with v=(t/x) Now you can use (dF/dx)=∫_0 ^1 (∂/∂x) g(x,v)dv where g(x,v)=xf(x−xv)cos(2xv)](https://www.tinkutara.com/question/Q80325.png)
Commented by mr W last updated on 02/Feb/20

Commented by mr W last updated on 02/Feb/20
