What is the Integral of Xsqrt(4+x^2) and the Integral of Xsqrt(4-x^2) ? Using trig substitution

Question

Using the techniques of trig substitution : ∫×√(4+x² )    dx and  ∫×√4-x² )

dx How can I solve these using trig substitution?

Answer 1

∫x√(4+x²)dx      Let x = 2tanθ    Then dx = 2sec²θdθ

 

So, the integral can be rewritten as ∫(2tanθ)√[4(1+ tan²θ)](2sec²θ)dθ

 

      = 8∫tanθsec³θdθ

 

      = 8∫sec²θ(tanθsecθ)dθ      Let u = secθ   Then du = secθtanθdθ

 

      = 8∫u²du = 8(1/3)u³ + C

 

                     = (8/3)sec³θ + C    Since x = 2tanθ, tanθ = x/2

 

                                                  Draw a right triangle with acute angle θ,

                                                  side opposite θ of length x, side adjacent

                                                  to θ of length 2, and hypotenuse √(4+x²).

 

So, (8/3)sec³θ + C = (8/3)[√(4+x²)/2]³ + C

 

                             = (1/3)(4+x²)^3/2 + C

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

∫x√(4-x²)dx    Let x = 2sinθ   Then dx = 2cosθdθ

 

So we get ∫(2sinθ)√[4(1-sin²θ)](2cosθ)dθ

 

             = 8∫sinθcos²θdθ

 

                        Let u = cosθ  Then du = -sinθdθ

 

                                                   -du = sinθdθ

 

So, 8∫sinθcos²θdθ = -8∫u²du = -(8/3)u³+C

 

                                            = -(8/3)cos³θ + C

 

         Since x = 2sinθ, sinθ = x/2

 

                   Draw a right triangle with acute angle θ, side opposite

                   angle θ of length x, hypotenuse 2, and side adjacent to θ

                   of length √(4-x²).

 

So, -(8/3)cos³θ + C = -(8/3)[√(4-x²)/2]³ + C

 

                              = -(1/3)(4-x²)^3/2 + C

 

Answer 2

∫x√(4 + x²)dx

 

Let        x = 2tanA                 x² = 4tan²A

           dx = 2sec²AdA              where  A is the angle

 

Substituting,

 

4∫tanA√(4 + 4tan²A)sec²A dA =

 

4∫tanA√(4(1 + tan²A)) sec²A dA =

 

8∫tanAsec³A dA =

 

8∫(sinA/cosA)(1/cos³A) dA

 

8∫(sinA / cos⁴A)dA

 

 

u = cosA

du = -sinAdA                 -du = sinAdA 

 

-8∫u^-4du =

 

-8 * (-1/3)u^-3 =

 

(8/3)u^-3 =

 

8 / (3u³)

 

 

Now we substitute into A.

 

8 / (3cos³A)

 

 

Substitute back into x.

 

8 / [3cos³ (arctan(x/2))] + C

 

 

 

For the second one, use the substitution    x = 4sinA.  Then apply trig identities afterwards.