23 Use a trigonometric substitution to evaluate the integra

23 - Use a trigonometric substitution to evaluate the integral

24- Integrate the function

Use a trigonometric substitution to evaluate the integral. Integrate the function. y2 / (25 - y2)3 / 2 dy

Solution

23)
x = 4sect
dx = 4secttantdt
I = 4secttantdt/4sect(16sec^2t - 16) dt
I = tant/tant dt
I = dt
integrating
I = t + C
I = arcsec(x/4) + C
24)
let y = 5sint
dy = 5costdt
I = 5cost*25sin^2t/(25-25sin^2t)^3/2dt
I = 125cost sin^2t/125cos^2t * cost
I = tan^2tdt
I = (sec^2t - 1)dt
integrating
I = tant - t + C
I = tan(sin^-1(y/5) - sin^-1(y/5) + C
I = tan(tan^-1(y/25-y^2) - sin^-1(y/5) + C
I = y/25-y^2- sin^-1(y/5) + C