To determine

To find:The value of the given expression.

Answer

limx→01x∫0x(1−tan2t)1tdt= e−2

Explanation

Given:

limx→01x∫0x(1−tan2t)1tdt

Formulae used: The fundamental theorem of calculus.Definition of derivative.

Consider limx→01x∫0x(1−tan2t)1tdt

Let (1−tan2t)1t=f(t) It is given that f is defined and continuous at t=0. Let the indefinite integral ∫(1−tan2t)1tdt=F(t)+c Then by fundamental theorem of calculus we have ∫0x(1−tan2t)1tdt=F(x)−F(0) and F'(t)=f(t) Thus limx→01x∫0x(1−tan2t)1tdt=limx→0F(x)−F(0)x By definition of derivative the expression on the RHS above is the derivative of F at zero. But F is anti-derivative of the integrand. Thus to find the required limit we only need to find the value of the integrand at zero. That is we need to find f(0).Summarising the discussion thus far we have limx→01x∫0x(1−tan2t)1tdt=limx→01x∫0xf(t)dt=limx→0F(x)−F(0)x=F'(0)=f(0) Since f is continuous at zero we have

f(0)=limt→0f(t)=limt→0(1−tan2t)1t=limt→0e1tln(1−tan2t)  [apply exponent ruleax=eln(ax)=ex⋅lna]

Simplify further and get:

limx→0e1tln(1−tan2x)=elimt→0[ln(1−tan2t)t]=elimt→0(−2sec2t1−tan2t1)  [apply L'hospital rule]=elimt→0−2cos2t(1−tan2t)=e−2cos2(0)(1−tan0)=e−2 Thus limx→01x∫0x(1−tan2t)1tdt=limx→01x∫0xf(t)dt=limx→0F(x)−F(0)x=F'(0)=f(0)=e−2

Conclusion: Hence, the value of the limit limx→01x∫01(1−tan2t)1tdt is e−2.