How do you integrate 1x216x7 using trigonometric substitution?

1 Answer
Jun 18, 2018

dxx216x7=ln(x8)+x216x7+C

Explanation:

Complete the square under the root:

dxx216x7=dx(x8)273

dxx216x7=173dx(x873)21

Substitute:

u=x873

du=dx73

so:

dxx216x7=duu21

Restrict now to the interval u(1,+) and substitute:

u=cosht

du=sinhtdt

with t(0,+)

dxx216x7=sinhtdtcosh2t1

As cosh2tsinh2t=1 and sinht>0 for t>0:

cosh2t1=sinht

dxx216x7=sinhtdtsinht=dt=t+C

undoing the substitution:

dxx216x7=arccosh(x873)+C

and using the logarithmic form of the inverse hyperbolic cosine:

dxx216x7=ln∣ ∣x873+(x873)21∣ ∣+C

Multiplying the argument of the logarithm by 73 only changes the constant C, so:

dxx216x7=ln(x8)+(x8)273+C

dxx216x7=ln(x8)+x216x7+C

and by differentiating we can verify that the solution is valid also for x8<73