Integrating Trig: Solving Definite Integrals with Sin and Cos

[MathGnome]

Ok, so we have

\int_{0}^{1}\left(\sin{2x}*\cos{2x}\right)dx

Using the double angle forumla we change the integrand

(1/2)\int_{0}^{1}\left(2*\sin{2x}*\cos{2x}\right)dx

which converts to

(1/2)\int_{0}^{1}\left(\sin{4x}\right)dx

This is where I run into trouble... I'm trying to use the formula
\int\left(\sin{bx}\right)dx = (1/-b)\cos{bx}

but my answers are not working. I'm thinking that it has something to do with the fact this is a deffinite integral... any help?

Thx

 

[amcavoy]

Ok, so we have

\int_{0}^{1}\left(\sin{2x}*\cos{2x}\right)dx

Using the double angle forumla we change the integrand

(1/2)\int_{0}^{1}\left(2*\sin{2x}*\cos{2x}\right)dx

which converts to

(1/2)\int_{0}^{1}\left(\sin{4x}\right)dx

This is where I run into trouble... I'm trying to use the formula
\int\left(\sin{bx}\right)dx = (1/-b)\cos{bx}

but my answers are not working. I'm thinking that it has something to do with the fact this is a deffinite integral... any help?

Thx

How are you answers not working?

-\frac{1}{4}\cos{4x}\bigvert|_0^1

As long as you know the value of cos(0), you shouldn't have any trouble!

 

[MathGnome]

... *sigh* I was so used to seeing a 0 and just subtracting by 0 that I forgot cos(0) is in fact NOT 0... Is it bad that I take shortcuts without even realizing I'm taking them?

EDIT: I meant WRONG shortcuts =)

 

[HallsofIvy]

By the way, I would not have worried about combining those trig functions.

To integrate \int_0^1 sin(2x)cos(2x)dx just note that the sine and cosine are both to an odd exponent (1). Let u= sin(2x) so that du= 2 cos(2x)dx or (1/2)du= cos(2x)dx. When x= 0, u= sin(0)= 0, when x= 1, u= sin(2). The integral becomes
\frac{1}{2}\int_0^{sin(2)}udu
which equals
\frac{1}{4}u^2 evaluated between 0 and sin(2) and is
\frac{1}{4}sin^2(2)