Answer :

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 =

∵ sin^{2}x = 1 – cos^{2}x = (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-x^{2}), 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-x^{2}) we get:

I = …(1)

Let I_{1} =

As derivative of 1 – x^{2} = -2x

∴ I_{1} =

The formula to be used:

&

∴ I_{1} =

Let 1 – x^{2} = u

⇒ du = -2x dx

I_{1} =

Thus,

I_{1} = -3√(1 - x^{2}) – 2 sin^{-1}x

∴ from equation 1:

I =

⇒ I =

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