# Evaluate the inte

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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