# How to integrate the following integral using trigonometric substitution? indefinite intergal x^2 /sqr ( 4 -x^2) dx

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### 1 Answer

Int x^2/ sqrt(4-x^2) dx

To use trigonometric substitution we follow the following identities:

We know that:

sqrt(a^2- x^2) ==> x = a*sin(t).............(1)

sqr(a^2+x^2) ==> x = a*tan(t)...............(2)

sqrt(x^2-a^2) ==> x = a*sec(t)................(3)

Comparing the given problem, we can apply the identity number (1).

==> sqrt(4-x^2) ==> x = 2*sin(t) ==> dx = 2cos(t) dt

==> x^2 = 4sin^2 (t)

Now we will substitute:

==> Int (4sin^2 (t)/sqrt(4-4sin^2 t) *2cos(t) dt

==> Int (8sin^2 t*cos(t) /sqrt(4(1-sin^2 t) dt

==> 8 Int (sin^2 t cos(t) / 2sqrt(cos^2 t) dt

==> 8 Int sin^2 t cos (t) / 2cos(t) dt

==> 4 Int sin^2 t dt

Now we know that sin^2 t = (1-cos2t)/2

==> 4 Int (1-cos2t)/2 dt

==> 2 Int (1- cos2t) dt

==> 2 [ t - sin2t/2 ] + C

==> 2t - sin2t + C

Now we will substitute with x = 2sint ==> sint = x/2 ==> t= arcsin(x/2)

==> 2(arcsin(x/2) - sin2t + C...........(i)

Since x= 2sint

sin2t = 2sint *cost ==> sin2t = x*cost

cost = sqr(1-sin^2t) = sqrt(1- (x/2)^2) = sqrt(1-x^2/4)= sqrt(4-x^2)/4 = (1/2)sqrt(4-x^2)

==> sin2t = 2sint*cost = x*(1/2)*sqrt(4-x^2)

==> sin2t = (1/2)*x*sqrt(4-x^2)

Substitute into (i) :

**==> Int x^2/sqrt(4-x^2) = 2arcsin(x/2) -(1/2)x*sqrt(4-x^2) + C**