Trigonometric Substitution Question

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January 21st 2010, 08:24 PM
Troopa

Trigonometric Substitution Question

Hello,

I am absolutely stumped on this one.

http://www.mario-kart.net/gamespy/im...Untitled-3.gif

I'm assuming all of my work is correct up until the end, but I don't know how to proceed using Trigonometric Substitution. Please help!
January 21st 2010, 08:37 PM
Jhevon

Quote:

Originally Posted by Troopa

Hello,

I am absolutely stumped on this one.

http://www.mario-kart.net/gamespy/im...Untitled-3.gif

I'm assuming all of my work is correct up until the end, but I don't know how to proceed using Trigonometric Substitution. Please help!

$\frac 1{\sec^2 \theta} = \cos^2 \theta = \frac 12 (1 + \cos 2 \theta)$

(and of course it's not $\frac 1{\tan \theta + C}$ there is no integration rule to justify such a step!)
January 21st 2010, 08:52 PM
Troopa

So then...

$= \frac 12 (1 + \cos 2 \theta)$

$= \frac 12 (\theta + (cos2 \theta^2 / 2))$

Is this right?
January 21st 2010, 08:54 PM
Jhevon

Quote:

Originally Posted by Troopa

So then...

$= \frac 12 (1 + \cos 2 \theta)$

$= \frac 12 (\theta + (cos2 \theta^2 / 2))$

Is this right?

no.

first, where's your integral sign?

secondly, how do we integrate cosine? We don't just simply integrate the argument.
January 21st 2010, 09:03 PM
VonNemo19

$\int\frac{1}{\sec^2\theta}d\theta=\int\cos^2\theta {d}\theta=\frac{1}{2}\int(1+\cos2\theta)d\theta=\f rac{1}{2}(\theta+\frac{1}{2}\sin{2\theta})+C$

Now back substitute.
January 21st 2010, 09:06 PM
Troopa

Thanks for your help. I'm gonna finish this and get some help in my math lab.

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