# Find: <img

Let, I = I = { sin 2x = 2 sin x cos x}

I = If sin x would have been present in numerator, we could have applied substitution method.

So let’s multiply sin x in numerator and denominator.

I = sin2x = 1 – cos2x = (1-cos x)(1+cos x)

I = Let cos x = t dt = -sin x dx

I = Now, this form clearly indicates that we need to apply the partial fraction method of integration.

So,

Let, -1 = A(1+t)(1+2t) + B(1-t)(1+2t) + C(1-t)(1+t)

Put t = 1,

-1 = A(1+1)(1+2)

6A = -1 or A = -1/6

Similarly on putting t = -1 we get

B = 1/2

And on putting t = -1/2 we get : C = -4/3 I = We know that: I = I = Put back t = cos x

I = OR

Let I = As we can observe that to apply the substitution method we only need linear terms, so let's try to separate them out. We also know the integration of √(1-x2), so we will try to get it.

Keeping this in mind add and subtract 1 in the numerator.

I = I = I = I = Formula to be used: Applying the formula for √(1-x2) we get:

I = …(1)

Let I1 = As derivative of 1 – x2 = -2x

I1 = The formula to be used: & I1 = Let 1 – x2 = u

du = -2x dx

I1 = Thus,

I1 = -3√(1 - x2) – 2 sin-1x

from equation 1:

I = I = Rate this question :

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