Question

How do you integrate `int 2/(y^4sqrt(y^2-25)` using trig substitutions?

Answer 1

`(2sqrt(y^2 - 25))/(625y) - (2(y^2-25)^(3/2))/(1875y^3) + C`

Explanation:

Our goal here is to get rid of the `√` by making use of a pythagorean identity. Let `y = 5sectheta`. Then `dy = 5secthetatanthetad theta`.

`=int 2/((5sectheta)^4sqrt((5sectheta)^2 - 25)) * 5secthetatanthetad theta`

Apply the identity `sec^2theta - 1 = tan^2theta`:

`=int 2/(625sec^4thetasqrt(25tan^2theta))* 5secthetatantheta d theta`

`=int 2/(625sec^4(5tantheta)) * 5secthetatantheta d theta`

`=int 2/(625sec^3theta) d theta`

Rewrite using `cosx = 1/secx`

`=int 2/625cos^3theta d theta`

`=int 2/625 cos^2theta(costheta) d theta`

Rearrange using the pythagorean identity `cos^2x + sin^2x = 1`:

`=int 2/625 (1 - sin^2theta)costheta d theta`

Let `u = sintheta`. Then `du = costheta d theta` and `d theta = (du)/costheta`.

`=int 2/625 (1 - u^2)costheta * (du)/costheta`

`=int 2/625( 1 - u^2) du`

Integrate using `intx^ndx = x^(n + 1)/(n + 1)` where `n != -1`.

`=2/625(u - 1/3u^3) + C`

`=2/625sintheta - 2/1875sin^3theta + C`

Draw a triangle to represent `y/5 = sectheta`.

`=2/625sqrt(y^2 - 25)/y - 2/1875(sqrt(y^2 - 25)/y)^3 + C`

`=(2sqrt(y^2 - 25))/(625y) - (2(y^2-25)^(3/2))/(1875y^3) + C`

Hopefully this helps!