Evaluate: xtan^2(x)dx from 0 to pi/3. Please i need help

To calculate integral of xtan^2x.

We now that tan^2x = sec^2x-1.

Therefore Int xtan^2x dx = Int x*(sec^2x - 1)dx

Int xtan^2x dx= Int xsec2xdx - Int xdx

Int xtan^2x dx = Intx*sec^2xdx - x^2/2....(1)

Int xsec^2x dx = xtanx - Int{x'tanx}dx.

Int xtanx dx =  xtanx - Int tanx dx

Int xtanx = xtanx+log(cosx) + C. Substituting this in (1) we get:

Int x tan^2x dx = xtanx + log(cosx) - x^2/2+C.

Now put limit from 0 to pi/3

(Pi/3)* sqrt(3)+ log(1/2) - pi²/18

Regards

Arun (askIITians forum expert)