Question

How do you integrate `int (4x)/sqrt(x^2-14x+45)dx` using trigonometric substitution?

Answer 1

`4sqrt(((x-7)/2)^2+1)+7cosh^-1((x-7)/2)+C`

Explanation:

We can begin by taking `x^2-14x+45` an re writing it in in completed square/ vertex form giving us:

`(x-7)^2-4` Putting this into the integral will give:

`int(4x)/sqrt((x-7)^2-4)dx`

We can factor the `4` out to the front then play with the numerator a little and we get:

`=4int(x-7+7)/sqrt((x-7)^2-4)dx`
`=4int(x-7)/sqrt((x-7)^2-4)+7/sqrt((x-7)^2-4)dx`

Now lets consider the substitution (we'll have to use a hyperbolic function, not a trig function): `2cosh(u)=x-7`
`2sinh(u)du=dx`

Now putting this substitution in we get:

`4int((2cosh(u))/sqrt(4cosh^2(u)-4)+7/sqrt(4cosh^2(u)-4))sinh(u)du`

Simplify this a little by factoring the `4` from the square root:

`=4int(sinh(u)cosh(u))/sqrt(cosh^2(u)-1)+(7sinh(u))/(2sqrt(cosh^2(u)-1))du`

Now at this point we can use the identity: `cosh^2(u)-1=sinh^2(u)` and then do a bit of cancelling.

That will give us:

`=4int(sinh(u)cosh(u))/sqrt(sinh^2(u))+(7sinh(u))/(2sqrt(sinh^2(u)))du`

`=4int(sinh(u)cosh(u))/sinh(u)+(7sinh(u))/(2sinh(u))du`

`=4intcosh(u)+7/2du`

Now evaluate the integral:

`4(sinh(u)+7/2u)+C=4sinh(u)+14u+C`

Now we can use the identity that we used earlier to convert `sinh` into `cosh`:

`=4(sqrt(cosh^2(u)+1)+7/2u)+C`

Now reverse the substitution and we get:

`4sqrt(((x-7)/2)^2+1)+7cosh^-1((x-7)/2)+C`

At this point you can re write the function in terms of its exponential and logarithmic definitions but I believe this maybe satisfactory>