Find the value of the Integral -p/2 to p/2 v(cos x-cos ³x)dx

taking cos x common from the term in the brackets the given integral sum is simplified to integral (vcosx).sinx dx substitute cosx = t(say) then -sinx.dx=dt the integral again becomes : integral(-pi/2 to pi/2) of -(vt).dt now we observe that the function root(cosx-cosx^3) is an even function as changing the variable x to its negative (-x) dosnt change the equation. so integral(-pi/2 to pi/2 ) becomes: 2*integral(0 to pi/2)-(vt).dt but wen x=0 cosx in 1. and wen x=pi/2 ,cosx is 0. so the integral again becomes: 2*integral(1 to 0) -(vt).dt now interchange the limts of the integral and take out the negative sign.the integral now is: 2*integral(0 to one)(vt).dt which is simply: 2*(2/3)*(t^3/2) from 0 to 1 which is -------->>>4/3.