Use substitution x=sin theta to evaluate integrate x^2/(1-x^2)^5/2

Question

Full working out please

Answer 1

I will denote angle theta by T.

 

x = sin T.   

Then Dx = cosT dT

 

Draw a right triangle so that the sine of angle T is x/1.

That is x is the opposite side of angle T and the hypotneuse is 1.

Then the adjacent side of angle T is sqrt(1 - x^2). So cos T = sqrt(1-x^2)

and tan T = x/sqrt(1-x^2). We will need these facts post-integration,

so keep them in mind.

 

The integral becomes:

 

 (sinT)^2 * cos T dT / (1 - (sinT)^2)^5/2 =

 

(sinT)^2 * cos T dT / ((cosT)^2)^5/2 =   <--- recall the basic trig identity sin^2 + cos^2 = 1;

                                                                        so 1 - sin^2 = cos^2

 

 

(sinT)^2 * cos T dT / (cosT)^5 =    <--- square root cancels the exponents of 2 ; that is,

                                                           the square of the square root is that same number

 

(sinT)^2 dT / (cosT)^4  =   <----   cosine in the numerator cancels one of the five cosines in the denominator

 

Now recall that tan = sin/cos and sec = 1/cos so the integral becomes:

 

(tanT)^2 * (secT)^2 dT =  

 

Let M = tanT

then dM = (secT)^2 dT.

 

The integral now becomes:

 

M^2 dM

 

which integrates to 1/3*M^3 + C  for some fixed number constant C

 

                1/3 * (tanT)^3 + C

 

               1/3(  x/sqrt(1-x^2) )^3 + C <--- from the identity derived above in bold based on the right triangle

                                                                     of sinT = x

 

 

             = 1/3(x^3/ ( 1 - x^2)^(3/2)  + c

 

             = 1/3(x^3)(1 - x^2)^(-3/2) + c  <--- this is the anti-derivative.

 

---------------------------------------------------------------------------

 

CHeck by differentiation:

 

Product Rule says:

       

  1/3*x^3  * (-3/2)*(1-x^2)^(-5/2) * (-2x) + 1/3(3x^2)(1 - x^2)^(-3/2)

 

Now cleaning up this derivative in raw form, the following algebra occurs:

 

    (first term, before the bolded plus sign)   .... 1/3 cancels the 3 in -3/2

                                                                .... negatives cancel as they multiply to +

                                                                .... the 2's cancel in -3/2 vs -2x

                                                                .... exponents of x combine to 4 so it becomes x^4

 

  (Second term, after the bolded plus sign) .... 1/3 cancels 3 leaving just x^2

 

 The derivative is now:

     (x^4)(1 - x^2)^(-5/2) + (x^2)(1 - x^2)^-3/2

 

Factoring out (1 - x^2)^(-5/2):

 

  (1 - x^2)^(-5/2)  [ x^4 + x^2(1 - x^2) ]   ;   recall:  -5/2 + 2/2  = -3/2

 

(1 - x^2)^(-5/2) [ x^4 + x^2 - x^4) ]  <--- distributes x^2 on the inside

 

(1 - x^2)^(-5/2) [ x^2] which is the original integrand.

Therefore the anti-derivative, the answer highlighted in bold and italic above, is correct and proven.

 

 

                                                           

                                           

Answer 2

Let   B = theta

 

 

x = sinB                   x² = sin²B

dx = cosB dB

 

 

Substituting,

 

sin²BcosB dB / (1 - sin²B)^5/2

 

 

Using trig identities,

 

sin²BcosB dB / (cos²B)^5/2 =

 

sin²BcosB dB / cos⁵B =

 

sin²B dB / cos⁴B =

 

(1 - cos²B) dB / cos⁴B =

 

(sec⁴B - sec²B) dB =

 

sec²B(sec²B - 1) dB =

 

sec²Btan²B dB

 

 

Let

r = tanB               dr = sec²BdB

 

 

Integral is now easy to evaluate.

 

∫r²dr = (1/3)r³

        

        = (1/3)tan³B

 

        = (1/3)tan³(arcsinx)