Integration Worksheet6 solution6

In this page integration worksheet6 solution6 we are going to see solution of some practice question from the worksheet of integration.

Question 14

Integrate the following with respect to x,   tan⁻¹ [(3 x - x³)/(1 - 3 x²)]

Solution:

Here we are going to apply the substitution method to integrate this problem

x = tan θ

Now we are going to apply x by tan θ in the given question,So tan⁻¹[(3x-x³)/(1-3x²)] will become

                  = tan⁻¹[(3tanθ - (tanθ)³)/(1-3(tan θ)²)]

                  = tan⁻¹[(3tanθ - tan³θ)/(1-3tan²θ)]

                  = tan⁻¹(tan 3 θ)

                  = 3 θ

By using the substitution method we have changed the original question as a simple term 3 θ. From this we come to know that integrating the given question is similar of integrating 3 θ. now we are going to replace θ by tan⁻¹ x

                y = 3 ∫ tan⁻¹ x dx

now w are going to apply the substitution method to integrate this

      u = tan⁻¹ x           dv = dx

   du = 1/(1+x²)           v = x

∫ u dv = uv - ∫ v du

         = (tan⁻¹ x) x -  ∫ x [1/(1+x²)] dx

         = (tan⁻¹ x) x -  ∫ x/(1+x²) dx

   t = 1 + x²

dt = 2 x dx

x dx = dt/2

         = (tan⁻¹ x) x -  ∫ (dt/2)(1/t)

         = (tan⁻¹ x) x -  (1/2)∫ 1/t dt     

         = (tan⁻¹ x) x -  (1/2) log t + C

         = (tan⁻¹ x) x -  (1/2) log (1+x²) + C


Question 15

Integrate the following with respect to x,   x sin⁻¹ (x²)

Solution:

Here we are going to apply the substitution method to integrate this problem

x² = t

2 x dx = dt

 x dx = dt/2

                  = ∫ x sin⁻¹ (x²) dx

                  = ∫ sin⁻¹ t (dt/2)

                  = (1/2) ∫ sin⁻¹ t dt

now we are going to apply the substitution method

u = sin⁻¹ t              dv = dt

du = 1/√(1-t²)         v = t 

∫ u dv = u v - ∫ v du

         = (1/2) [(sin⁻¹ t)t -∫ t (1/√(1-t²))dt]

         = (1/2) [t(sin⁻¹ t) -∫ t/√(1-t²)dt] ------ (1)

now let us find the integration value of ∫ t/√(1-t²) dt after that we can apply the value in the first equation.

let a = 1 - t²

    da = - 2 t dt

    -da/2 = t dt

∫ t/√(1-t²) dt = ∫ (-da/2)/√a

                    = - (1/2)∫ (1/√a) da

                    = - (1/2) [√a/(1/2)]

                    = - (1/2) [2√a]

                    = - √(1-t²) + C

now we are going to apply this in the first equation

         = (1/2) [x²(sin⁻¹ x²) - (- √(1-t²))]+ C

         = (1/2) [x²(sin⁻¹ x²) + √(1-(x²)²)]+ C

         = (1/2) [x²(sin⁻¹ x²) + √(1-x⁴)]+ C

integration worksheet6 solution6 integration worksheet6 solution6