Question

How do you integrate `int 1/sqrt(x^2-9x+8)` using trigonometric substitution?

Answer 1

`I=int1/sqrt(x^2-9x+8)dx=int1/sqrt((x-9/2)^2-(7/2)^2)dx`
We know that, `int1/sqrt(X^2-A^2)dX=ln|X+sqrt(X^2-A^2)|+c`
Put,`X=x-9/2,andA=7/2`,then simplify,term in sq.rt.
`I=ln|(x-9/2)+sqrt(x^2-9x+8)|+C`

Explanation:

Without Trig.Subst. answer ,see above.

With Trig. Subst. answer, see below.

Decide yourself ,which method is better.

We have, `x^2-9x+color(red)8=x^2-9x+color(red)(32/4)=x^2- 9x+color(red)(81/4-49/4)`

`:.x^2-9x+8=(x-9/2)^2-(7/2)^2`

#i.e.I=int1/sqrt((x-9/2)^2-(7/2)^2)dx#

Let, `x-9/2=7/2secu=>dx=7/2secutanudu`

So,

`I=int(7/2secutanu)/sqrt((7/2secu)^2-(7/2)^2)du`

`=int(7/2secutanu)/(7/2sqrt(tan^2u))du`

`=int(secutanu)/tanudu`

`=intsecudu`

`=ln|secu+tanu|+c`

`=ln|secu+sqrt(sec^2u-1)|+c,to[where,secu=(x-9/2)/(7/2)]`

`=ln|(x-9/2)/(7/2)+sqrt(((x-9/2)/(7/2))^2-1)|+c`

`=ln|(x-9/2)/(7/2)+sqrt((x-9/2)^2-(7/2)^2)/(7/2)|+c`

`=ln|(x-9/2)/(7/2)+sqrt(x^2-9x+8)/(7/2)|+c`

`=ln|(x-9/2+sqrt(x^2-9x+8))/(7/2)|+c`

`=ln|x-9/2+sqrt(x^2-9x+8)|-ln(7/2)+c`

`=ln|x-9/2+sqrt(x^2-9x+8)|+C,where,C=c-ln(7/2)`

Answer 2

`=> cosh^(-1) ((2x-9)/7 ) + c`

Explanation:

`=> int 1/ ( sqrt(( x-9/2)^2 -49/4 )`

`=> u = (2x-9) / 7`

`=> dx = 7/2 du`

`=> int 7 / sqrt( 49u^2 - 49 ) dx`

`=> int 1/ sqrt(u^2 -1 ) du`

`=> u = cosh theta`

`=> du = sinh theta d theta`

`=> int (sinh theta d theta ) /sqrt(cosh^2 theta -1)`

`=> int d theta`

`=> theta + c`

`=> cosh ^(-1)u + c`

`=> cosh^(-1) ((2x-9)/7 ) + c`