since i'm once again bored, I thought I'd share a problem with you.

 

Solve :int: sqrt(1+sin(2x))dx

 

I've seen a very nice way of doing this that made me kick myself after I'd seen it, so i'll let you all have a look at it before before spoiling the fun

 

have fun.

Just use :int:uv = uv - :int: v du

 

u=1+sin(2x)

v=sqrt(1+sin(2x))

 

du=2cos(2x) (chain rule)

 

(1+sin(2x))(sqrt(1+sin(2x))) - :int: (sqrt(1+sin(2x)))(2cos(2x))

 

(sin(2x)+1)^(3/2) - :int: (cos(2x)/sqrt(sin(2x)+1)

 

(sin(2x)+1)^(3/2) - sqrt(sin(2x)+1)

 

which you can further simplify to (sin(2x))(sqrt(sin(2x)+1))

 

 

 

i'm sure i made a stupid mistake somewhere, but the general idea is right... and for some reason my calculator can't do this problem :/

Attempt number two:

 

(sin(2x)+1)^(3/2) - :int: 2sqrt(sin(2x)+1)*cos(2x)

 

(sin(2x)+1)^(3/2) - (2(sin(2x)+1))^(3/2))/3

 

=== (sin(2x)+1)^(3/2))/3

 

 

There's still a stupid mistake, but this is closer. It usually takes me 5-6 attempts to solve a problem without stupid mistakes.

Dude,

 

If you're bored then please take a look at my differential equation problem.

 

I'm having trouble finding a good resourse on the net that lays out an introduction to differential equations and is full of worked problems. If you know of one, please share.

 

Thank You!

nice method:

 

you can use the fact that (sin(x))^2 + (cos(x))^2 = 1 to solve this problem.

 

clearly this implies that:

 

:int: sqrt((sin(x))^2 + sin(2x) + (cos(x))^2)dx

= :int: sqrt((sin(x))^2 + 2sin(x)cos(x) + (cos(x))^2)dx

 

but (sin(x))^2 + 2sin(x)cos(x) + (cos(x))^2 = (sin(x)+cos(x))^2, so the integral boils down to:

 

:int: (sin(x) + cos(x))dx

 

which i thought was rather nifty

 

in regard to the differential equation problem, i have looked at it, but it is (as you say) a bit of a toughie. i may have a look at it in a bit.

Trig identities are evil.

sin(x) + cos(x) :neq: sqrt(1+sin(2x))

if you want to be pedantic, then |sin(x) + cos(x)| = sqrt(1+sin(2x))

 

but even so, i thought it was a pretty method for solving a problem that looks fairly nasty.

nice lil identity