Q. 214.5( 2 Votes )

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Answer :

We have to find the integral of but first we will convert this trigonometric function into algebraic function by substitution which is,

Put ex = t, so,


= ex,


Hence ex dx = dt, putting in the function we get,


I =


We will solve this function by using partial fractions.



A(t + 2)(t - 1) + B(t + 2) + C(t - 1)2 = t01 + t10 + t20


At2 + At - 2A + Bt + 2B + Ct2 + C – Ct = t01 + t10 + t20


A + C = 0, A + B – C = 0, –2A + 2B + C = 1


B = 2C, 2C + 4C + C = 1,


C = , A = , B =


Now making a function into partial fractions, we get,


I =


I = ln ln + C


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