what is the integration of xtanx?

Question

integration of xtanx

Answer 1

If we integrate by parts, where u=x (du=dx) and dv=tan(x)dx (v=ln|sec(x)|, we get:

xln|sec(x)|+∫ln|cos(x)|dx.

The integral here can only be evaluated as a series (Taylor/Maclaurin).

Let y(x)=∑a_nxⁿ=ln|cos(x)|, then y(0)=a₀=ln|cos(0)|=0; let t=tan(x) for convenience and brevity. When x=0, t=0. The summation starts at n=0.

Whenever we differentiate t wrt x, we get 1+t² in the derivative.

The rth derivative of y is represented by y^(r), so y'=dy/dx=y⁽¹⁾.

y⁽¹⁾(x)=∑na_nx^n-1, for n≥1, so the first term is constant a₁;  y⁽¹⁾(x)=-tan(x)=-t; y⁽¹⁾(0)=0, making a₁=0. 

With each derivative n≥r for the rth derivative. The first term (constant) is r!a_r. All other terms vanish when x=0.

y⁽²⁾(x)=∑n(n-1)a_nx^n-2=-sec²(x)=-(t²+1); y⁽²⁾(0)=2!a₂=-1, a₂=-½.

y⁽³⁾(x)=∑n(n-1)(n-2)a_nx^n-3=-2sec²(x)tan(x)=-2t(t²+1)=-2t³-2t; y⁽³⁾(0)=3!a₃=0.

y⁽⁴⁾(x)=-6t²(t²+1)-2(t²+1)=-6t⁴-8t²-2, a₄=-2/4!=-1/12.

y⁽⁵⁾(x)=-(24t³+16t)(t²+1)=-24t⁵-40t³-16t, a₅=0.

y⁽⁶⁾(x)=-(120t⁴+120t²+16)(t²+1)=-120t⁶-240t⁴-136t²-16, a₆=-16/6!=-1/45.

a_2k+1=0. Further derivatives and coefficients a_2k can be found.

However, it remains to perform integration of this series.

∫∑a_nxⁿdx=∑a_nxⁿ⁺¹/(n+1). a₀=a_2k+1=0, a₂=-½, a₄=-1/12, a₆=-1/45.

∫ln|cos(x)|dx=-x³/6-x⁵/60-x⁷/315-...

∫xtan(x)dx=xln|sec(x)|-x³/6-x⁵/60-x⁷/315-...+C.