Q. 454.1( 11 Votes )

Evaluate the inte

Answer :

I = =


I =


Let, sin x = t cos x dx = dt


I =


As we can see that there is a term of t in numerator and derivative of t2 is also 2t. So there is a chance that we can make substitution for t2 – 4t + 4 and I can be reduced to a fundamental integration.


As,


Let, 3t – 2 = A(2t – 4) + B


3t – 2 = 2At – 4A + B


On comparing both sides –


We have,


2A = 3 A = 3/2


–4A + B = –2 B = 4A – 2 = 4


Hence,


I =


I =


Let, I1 = and I2 =


Now, I = I1 + I2 ….eqn 1


We will solve I1 and I2 individually.


As, I1 =


Let u = t2 – 4t + 4 du = (2t – 4)dx


I1 reduces to


Hence,


I1 = { }


On substituting value of u, we have:


I1 =


I1 = ….eqn 2


I2 =


I2 =


Using: a2 – 2ab + b2 = (a – b)2


We have:


I2 =


As,


I2 = + C …eqn 3


From eqn 1, we have:


I = I1 + I2


Using eqn 2 and 3, we get –


I =


Putting value of t in I:


I = …..ans

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