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

1. May 12, 2012

### 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!!

2. May 12, 2012

### yus310

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

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

3. May 12, 2012

### HallsofIvy

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

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

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.

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

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