Integral of hellalot of work is shown

[johnq2k7]

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

 

[Mark44]

The first thing you should do is break this into two integrals.
$$2\pi \int_0^{\sqrt{2}} (2 - y^2)\sqrt{1 + 4y^2}dy$$
$$= 2\pi \int_0^{\sqrt{2}} 2\sqrt{1 + 4y^2}dy - 2\pi \int_0^{\sqrt{2}} y^2 \sqrt{1 + 4y^2}dy$$

I think that both can be tackled with the same trig substitution, namely tan u = 2y.