Evaluate the integral (help) sin2x dx b=pi/8 a=0

[needmathhelp!]

Evaluate the integral (help)! sin2x dx b=pi/8 a=0

∫sin2x dx b=pi/8 a=0

I'm having a little trouble finishing out the problem. This is what I've got so far.

u=2x
du=2dx
du= 1/2dx

1/2∫sin(u) du
(-1/2)cos u
(-1/2) cos 2x

[(-1/2)cos2(pi/8)]-[(-1/2)cos(2(0))]
? + (1/2)


So... The homework question gave me the answer already, but I do not understand how the answer is ... (2 - sqrt(2) / 4)

Can someone please "spell it out for me"...
Thanks for your help!

 

[yus310]

∫sin2x dx b=pi/8 a=0

I'm having a little trouble finishing out the problem. This is what I've got so far.

u=2x
du=2dx
du= 1/2dx

1/2∫sin(u) du
(-1/2)cos u
(-1/2) cos 2x

[(-1/2)cos2(pi/8)]-[(-1/2)cos(2(0))]
? + (1/2)


So... The homework question gave me the answer already, but I do not understand how the answer is ... (2 - sqrt(2) / 4)

Can someone please "spell it out for me"...
Thanks for your help!

You are asking the wrong question.

Ok, so what is the equivalent to sin(2x), in terms of sin(x), cos(x), equally,

WHY

u=2x
HOW DID YOU GO FROM

du=2dx

to
du= 1/2dx?

and why? It seems this is the problem, if you take a look.

Cheers,
yus310

 

[HallsofIvy]

First, this is a question about Calculus, not "differential equations" so I will move it.

∫sin2x dx b=pi/8 a=0

I'm having a little trouble finishing out the problem. This is what I've got so far.

u=2x
du=2dx
du= 1/2dx

I assume this was a typo since you don't continue it. If du= 2x, then dx= (1/2)du, not "du= (1/2)dx" as you have.

1/2∫sin(u) du

But here you have replaced dx with (1/2) du so you have done it correctly.

(-1/2)cos u
(-1/2) cos 2x

[(-1/2)cos2(pi/8)]-[(-1/2)cos(2(0))]
? + (1/2)

$2(\pi/8)= \pi/4$. Surely you know that $cos(\pi/4)= \sqrt{2}/2$?

So... The homework question gave me the answer already, but I do not understand how the answer is ... (2 - sqrt(2) / 4)

Can someone please "spell it out for me"...
Thanks for your help!