How do you integrate x216x2 by trigonometric substitution?

2 Answers
Jun 12, 2018

8sin1(x4)x16x22+C

Explanation:

I=x216x2dx

Let's use the substitution x=4sinθ. This substitution implies that couple important things:

  • x2=16sin2θ
  • 16x2=1616sin2θ=41sin2θ=4cosθ
  • dx=4cosθdθ

Now, let's substitute these back into our integral:

I=16sin2θ4cosθ(4cosθdθ)=16sin2θdθ

To solve this, I like to use the identity sin2θ=12(1cos2θ). If you've never seen this before, remember that cos2θ=cos2θsin2θ, which implies that cos2θ=12sin2θ, from which you can solve for sin2θ to get the identity from before. Using the identity in the integral:

I=1612(1cos2θ)dθ=8(θ12sin2θ)+C

We need to get our final answer back in terms of x instead of θ. Use sin2θ=2sinθcosθ:

I=8θ8sinθcosθ+C

Recall we started with x=4sinθ. Thus:

  • sinθ=x4
  • cosθ=1sin2θ=1x216=16x24
  • θ=sin1(x4)

And finally:

I=8sin1(x4)8(x4)(16x24)+C

I=8sin1(x4)x16x22+C

Jun 12, 2018

x216x2dx=8arcsin(x4)x16x22+C

Explanation:

Let x=4sint with t(π2,π2), dx=4costdt, then:

x216x2dx=(4sint)2(4costdt)1616sin2t

x216x2dx=16sin2tcostdt1sin2t

For t(π2,π2) the cosine is positive, so:

1sin2t=cos2t=cost

then:

x216x2dx=16sin2tcostdtcost=16sin2tdt

Use now the trigonometric identity:

2sin2α=1cos(2α)

to have:

x216x2dx=8(1cos2t)dt

and based on the linearity of the integral:

x216x2dx=8dt4cos2td(2t)

x216x2dx=8t4sin2t+C

x216x2dx=8t8sintcost+C

and undoing the substitution:

x216x2dx=8arcsin(x4)2x1(x4)2+C

x216x2dx=8arcsin(x4)x16x22+C