#### 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.