Question

How do you integrate `int 1/(usqrt(5-u^2))` by trigonometric substitution?

Answer 1

The answer is `-1/sqrt(5) ln|(sqrt(5) + sqrt(5 - u^2))/u| + C`

Explanation:

Use the substitution `u = sqrt(5)sintheta`. Then `du = sqrt(5)costhetad theta`. Calling the integral `I`, we have:

`I = int 1/(sqrt(5)sin thetasqrt(5 - (sqrt(5)sin theta)^2)) * sqrt(5)costheta d theta`

`I = int 1/(sqrt(5)sin theta sqrt(5 - 5sin^2theta)) * sqrt(5)costheta d theta`

We can now simplify the square root in the denominator.

Since: `sqrt(5 - 5sin^2theta) = sqrt(5(1 - sin^2theta)) = sqrt(5(cos^2theta)) = sqrt(5)costheta` because `sin^2theta + cos^2theta= 1`.

`I = int 1/(sqrt(5)sin theta sqrt(5)costheta) * sqrt(5)costheta d theta`

`I = int1/(sqrt(5)sin theta) d theta`

`I = int1/sqrt(5) csc theta d theta`

The integral of cosecant is known.

`color(blue)(int cscx dx= -ln|cscx + cotx|)`

Therefore:

`I = -1/sqrt(5)ln|csctheta + cottheta|`

From our initial substitution, we know that `sintheta= u/sqrt(5)` and therefore `csctheta = 1/sintheta = sqrt(5)/u`. Also, `cottheta =sqrt(5 - u^2)/u`.

`I = -1/sqrt(5)ln|sqrt(5)/u + sqrt(5 - u^2)/u|`

Simplify, and don't forget to add the constant of integration at the end!

`I = -1/sqrt(5)ln|(sqrt(5) + sqrt(5 - u^2))/u| + C`

Hopefully this helps!