- #1
johnq2k7
- 64
- 0
Evaluate the integral below:
2Pi times the integral of (2-y^2)(sqrt(1+4y^2)) dy from 0 to sqrt(2)
work shown:
using integration by parts I got:
let u= sqrt(1+4y^2)
therefore du= 4y/(4y^2+1)
let dv= (2-y^2)
therefore V= 2y- y^3/3
since Integration by Parts is (u)(V) - integral of (V)(du)
therefore i got 2*Pi times (sqrt(1+4y^2)(2y-y^3/3) - integral of (2y-y^3/3)(4y/sqrt(4y^2+1)) dy from 0 to sqrt (2)
using trig. subs. method i got
i tried using the tan y= 2y method of trigonometric substition
but i still wasn't able to evaluate the integral fully
i still need some help...
here's my work shown below:
2Pi times integral of (1-2y^2)*(sqrt(1+4y^2)) from 0 to sqrt(2) becomes:
2PI times integral of (1- ytan y)(sqrt(1+tan^2(y)) from 0 to sqrt(2)
since sec^2(y)= 1+tan^2(y)
i got 2Pi times integral of (1- ytan(y))(sqrt(sec^2(y))
therefore i got 2Pi times integral of (1- y*tan(y))(sec(y))
since sec y= 1/ cos y and tan x= sin y/ cos y
if i let u = sec y
du= sec y tan y dy
therefore i got, 2Pi times integral of u - y du from 0 to sqrt(2)
i still need a lot of help... I've tried using normal integration by parts methods.. and i haven't been able to fully evaluate the integrate because it keeps creating another integral that needs to evaluated by parts again.. .i need a lot of help here
after evaluating this i keep getting a more completed integration by parts.. and continuous iteration.. please help me solve this integral
2Pi times the integral of (2-y^2)(sqrt(1+4y^2)) dy from 0 to sqrt(2)
work shown:
using integration by parts I got:
let u= sqrt(1+4y^2)
therefore du= 4y/(4y^2+1)
let dv= (2-y^2)
therefore V= 2y- y^3/3
since Integration by Parts is (u)(V) - integral of (V)(du)
therefore i got 2*Pi times (sqrt(1+4y^2)(2y-y^3/3) - integral of (2y-y^3/3)(4y/sqrt(4y^2+1)) dy from 0 to sqrt (2)
using trig. subs. method i got
i tried using the tan y= 2y method of trigonometric substition
but i still wasn't able to evaluate the integral fully
i still need some help...
here's my work shown below:
2Pi times integral of (1-2y^2)*(sqrt(1+4y^2)) from 0 to sqrt(2) becomes:
2PI times integral of (1- ytan y)(sqrt(1+tan^2(y)) from 0 to sqrt(2)
since sec^2(y)= 1+tan^2(y)
i got 2Pi times integral of (1- ytan(y))(sqrt(sec^2(y))
therefore i got 2Pi times integral of (1- y*tan(y))(sec(y))
since sec y= 1/ cos y and tan x= sin y/ cos y
if i let u = sec y
du= sec y tan y dy
therefore i got, 2Pi times integral of u - y du from 0 to sqrt(2)
i still need a lot of help... I've tried using normal integration by parts methods.. and i haven't been able to fully evaluate the integrate because it keeps creating another integral that needs to evaluated by parts again.. .i need a lot of help here
after evaluating this i keep getting a more completed integration by parts.. and continuous iteration.. please help me solve this integral