How do you find the definite integral of #(1/x^2)*sin[(x-1)/x] dx# from #[1,2]#?

Question

5758 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-29T12:37:14

#1-cos (1/2 radian)=0.1224#

#dcos((x-1)/x)#

#=dcos(1-1/x)#

#=(cos(1-1/x))' dx#

#=-sin(1-1/x) (1-1/x)' dx#

#=-sin(1-1/x)( -(-1)1/xx^2) dx#

So, the integral is simply #int -(dcos^(-1)((x-1)/x)) #, between the limits x = 1 and x = 2.

#=-cos((x-1)/x)#, between the limits

#=-[ cos(1/2#-cos(0)]#

#=1-cos(1/2)#.

Answer 2 · 2016-09-29T14:05:47

#1-cos(1/2), or, 2sin^2(1/4)#.

Let #I=int_1^2 (1/x^2)sin[(x-1)/x]dx=int_1^2 (1/x^2)sin(x/x-1/x)dx#

#:. I=int_1^2 (1/x^2)sin(1-1/x)dx#

We subst., #(1-1/x)=t rArr {0-(-1/x^2)}dx=dt, or, 1/x^2dx=dt.#

Also, #x=1 rArr t=(1-1/x)=0, and, x=2 rArr t=1/2.#

Hence, #I=int_1^2 (1/x^2)sin(1-1/x)dx#

#=int_0^(1/2) sintdt#

#=[-cost]_0^(1/2)#

#=-[cos(1/2)-cos0]#

#:. I=1-cos(1/2), or, 2sin^2(1/4)#.

Enjoy Maths.!