integration help... work shown (2-y^2)(sqrt(1+4*y^2)

[djdavis2k]

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)

after evaluating this i keep getting a more completed integration by parts.. and continous iteration.. please help me solve this integral

since integral of

 

[Deleted member 4993]

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)

after evaluating this i keep getting a more completed integration by parts.. and continous iteration.. please help me solve this integral

since integral of

Have learned trigonometric substitutions. If you did, then try:

\(\displaystyle 2y \, = \, \tan(\theta)\)

 

[djdavis2k]

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)

after evaluating this i keep getting a more completed integration by parts.. and continous iteration.. please help me solve this integral

since integral of

Have learned trigonometric substitutions. If you did, then try:

\(\displaystyle 2y \, = \, \tan(\theta)\)

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

 

[Deleted member 4993]

i tried using the tan y= 2y method of trigonometric substition <<< That is not correct
I hope you meant

tan (x)= 2y

and

sec[sup:1pn5jnou]2[/sup:1pn5jnou](x) dx = 2 dy

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) << Where did that comefrom?

What happened to dy?

The limits need to change accordingly.

You need to review method of integration by substitution - thoroughly.
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