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 =


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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