Question

how to solve this?please help me

Answer 1

\[\int\limits_{0}^{1}y2\sqrt{4y2+1}\]

Answer 2

its y^2 right ?

Answer 3

yup

Answer 4

i see..modify it like this

y^2 (sqrt(4y^2 +1)) = (1/4)* (4y^2)(sqrt(4y^2 +1))
= (1/4)* (4y^2 +1 -1)(sqrt(4y^2 +1))
= (1/4)* (4y^2 +1)(sqrt(4y^2 +1)) - (1/4)*(sqrt(4y^2 +1))

= (1/4)* (4y^2 +1)^(3/2) - (1/4)*(sqrt(4y^2 +1))

now integrate both separately,,should be easier now..

Answer 5

for the first integral, express in the form of(y^2 +1/4)^3/2
now let y= 1/2 sinx or 1/2cos x ,,should help..

are you following ?

Answer 6

wait,i try to work on it
thnks ya

Answer 7

take your time..
'll be glad if i could help.. :)

Answer 8

ohh wait,,leme try for an alternative solution,

let y= 1/2 tanx ,,change limits accordingly..

=> after simplifying, and taking out all constants, inside the integral, you'll be left with

tan^2 x* secx

now this is 100 times easier than the first method! sorry..

Answer 9

is there any other solution? like using the u substitution?

Answer 10

u substitution ?

Answer 11

integration by substitution

Answer 12

2nd method which i stated seems to be easiest to me..
tan^2x = sec^2x - 1 ..you'll have to make use of this..

Answer 13

thats what i used! o.O

Answer 14

i never used this kind of method before. the trigonometric seem so complicated

Answer 15

i see,,then maybe we shall wait and hope anyone else would describe a better method.. even i'll get to learn then :)

Answer 16

i hope so

Answer 17

No trig sub? Hmmm ok, I think we can do integration by parts on this one.\[\large \int\limits_0^1 y^2\sqrt{4y^2+1}\;dy \quad = \quad \int\limits_0^1 \color{cadetblue}{y}\cdot \color{plum}{y \sqrt{4y^2+1}\; dy}\]\[\huge \color{cadetblue}{u=y}, \qquad \color{plum}{dv=y \sqrt{4y^2+1}\;dy}\]

Finding v is a bit tough, you might want to do a u-sub, or rather call it like an m-sub or something, since we're already using u in our by-parts.
Anyway, you should get something like this for du and v.\[\huge \color{cadetblue}{du=dy}, \qquad \color{plum}{v=\frac{1}{12}(4y^2+1)^{3/2}}\]

Answer 18

Which gives us,\[\large \frac{1}{12}y(4y^2+1)^{3/2}-\frac{1}{12}\int\limits(4y^2+1)^{3/2}dy\]Hmm that leaves us with a bad integral, need trig for this method also it seems :( That stinks...

Answer 19

yes,,still trigo would be required..your method was also quiet efficient though! :)

Answer 20

I just wanted a chance to use the colors XD lol
Just realized you can do colors in the equation tool, so i've been experimenting :3 hehe

Answer 21

i hardly used eqn tab in OS eversince i joined OS,, :D

am just to bad with latex,,its very time consuming for me unlike many here! :|