Q. 94.8( 5 Votes )

# 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 =

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