# Given the function f(x)=integral(0down-x up) (sint+cost)sint/(cost)^2,calculate limit f(x)/x^2 if x>0 0<x<pie/2

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The function `f(x) = int_0^x ((sin t + cos t)*sin t)/(cos^2t) dt` . The limit `lim_(x->0) (f(x))/x^2` has to be evaluated.

`int_0^x ((sin t + cos t)*sin t)/(cos^2t) dt`

= `int_0^x (sin^2 t + sin t*cos t)/(cos^2t) dt`

= `int_0^x tan^2 t + (sin(t))/(cos(t)) dt`

= `int_0^x sec^2 t - 1 dt + int_0^x (sin(t))/(cos(t)) dt`

`int_0^x 1 - sec^2 t dt`

= `[tan t - t]_0^x`

= tan x - x

`int_0^x (sin(t))/(cos(t)) dt`

let `cos(t) = y` , `dy/dt = -sin(t)`

`int (sin(t))/(cos(t)) dt = int -1/y dy = -ln(cos(t))`

`int_0^x (sin(t))/(cos(t)) dt = [-ln(cos(t))]_0^x = -ln(cos x)`

f(x) = `tan x - ln(cos x) - x`

Substituting x = 0 in limit `lim_(x->0) (f(x))/x^2` gives the indeterminate form `0/0` . Use l'Hopital's rule and substitute the numerator and denominator with their derivatives.

= `lim_(x->0) (sec^2x - tan x - 1)/(2x)`

= `lim_(x->0) (tan^2x - tan x)/(2x)`

= `0 - 1/2`

**The required limit is `-1/2` **