Question

If, $J=\underset{-5}{\overset{-4}{\int }}(3-x^2)\tan (3-x^2)dx$ and $K=\underset{-2}{\overset{-1}{\int }}(6-6x+x^2)\tan (6x-x^2-6)dx,$ then $(J+K)$ equals to

Answer 1

Given, $J=\underset{-5}{\overset{-4}{\int }}(3-x^2)\tan (3-x^2)dx$
Putting $(x+5)=t,$ we get
$J=\underset{0}{\overset{1}{\int }}(3-(t-5{)}^2\left)\tan \right(3-(t-5{)}^2)dt=\underset{0}{\overset{1}{\int }}(-22+10t-t^2)\tan (-22+10t-t^2)dt.$
Now, $K=\underset{-2}{\overset{-1}{\int }}(6-6x+x^2)\tan (6x-x^2-6)dx$
Putting $(x+2)=z,$ we get
$K=\underset{0}{\overset{1}{\int }}(6-6(z-2)+(z-2{)}^2\left)\tan \right(6(z-2)-(z-2{)}^2-6)dz=\underset{0}{\overset{1}{\int }}(22-10z+z^2)\tan (-22+10z-z^2)dz=-J$
Hence, $(J+K)=0$