Question

I need help in evaluating the integral of:

(sin(2x))/(cos^2(2x))dx, from x=0 to pi/6
I began evaluating this integral using u-substitution, with u = cos^2(2x). My big question, at this point is: what does "dx" equal? And, how does the integral look when everything is in terms of u?

Any and all help is greatly appreciated! :)

Answer 1

I've determined thus far..
u = cos^2(2x)
du = -2(sin(4x))dx

Answer 2

\[\int_0^{\pi/6}\frac{\sin 2x}{\cos^22x}~dx\]
Substitute \(u=\cos2x\), then \(-\dfrac{1}{2}du=\sin2x~dx\).
\[-\frac{1}{2}\int_1^{1/2}\frac{1}{u^2}~du\]

Answer 3

1. What happened to the limits?
2. So... I should use a different u?

Answer 4

1. It comes as a result of changing the variable. You're integrating with respect to \(u\) now. Before, \(x\) varied from 0 to \(\dfrac{\pi}{6}\), but we let \(u=\cos2x\). This means \(u\) will vary as follows:
\[x=0~~\Rightarrow~~u=\cos(2\times0)=\cos0=1\\
x=\frac{\pi}{6}~~\Rightarrow~~u=\cos\left(2\times\frac{\pi}{6}\right)=\cos\frac{\pi}{3}=\frac{1}{2}\]
2. Yes, the substitution you used doesn't help much.

Answer 5

To add to (1), you can keep the limits as they were, but you have to be careful. If you'd like to do so, you would have
\[\int_{x=0}^{x=\pi/6}\frac{1}{u^2}~du=\left[-\frac{1}{u}\right]_{x=0}^{x=\pi/6}=\left[-\frac{1}{\cos2x}\right]_{x=0}^{x=\pi/6}\]
I prefer my method, where you can just substitute the limits in directly for \(u\).

Answer 6

Okay. I'm going to start my problem over, from square one. Will you be online/ monitoring this question?
And, I often use radian limits (I love graphing calculators!) so I'm going to go with that route :)

Answer 7

Here's where I am:
u = cos(2x)
du = -2sin(2x)dx

Answer 8

(-1/2)(u^(-2))du