# trig substitution

• Feb 5th 2010, 11:55 AM
driver327
trig substitution
1. (integral) dx / sqrt(9+x^2)
I had a^2 = 9, a=3, x = 3tan(theta), dx = 3(sec(theta))^2, and
sqrt(9+x^2) = 3sec(theta)

I plugged them in and got:

3(sec(theta))^2 / 3(sec(theta)) = (integral) 3sec(theta).

I then integrated it and got ln|sec(theta) + tan(theta)| +c

I just divided the 3 and got: sqrt(9+x^2) / 3

and for tan(theta), i divided the 3 again and got tan(theta)=x/3

So, the ending result I had gotten was :

ln|(sqrt(9+x^2) / 3) + (x/3)| +c But the answer in the book had the same answer, but the 3's were not in. What am I doing wrong??

2.(3/2,0)(integral) dx/sqrt(9- x^2)

I used x = 3sin(theta), a = 3, dx = 3cos(theta), and

sqrt(9- x^2) = 3cos(theta)

I got the integral: 3cos(theta)/3cos(theta) = integral 1 (theta)

I then integrated it and just got theta.

I inversed the sin and got x = 3sin(theta)^-1

My problem is that I plugged the numbers in, and it was undefined. the

answer in the book came out to pi/6

plz help
• Feb 5th 2010, 12:26 PM
tom@ballooncalculus
1) With sqrt(1 + x^2) in the denominator you'll find sinh^-1 an easier sub.

Just in case a picture helps...

http://www.ballooncalculus.org/asy/maps/internal.png

... where

http://www.ballooncalculus.org/asy/chain.png

... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to x, and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

ln|(sqrt(9+x^2) / 3) + (x/3)|

= ln|(sqrt(9+x^2 + x) / 3)|

= ln|sqrt(9+x^2 + x)| - ln|3|

and ln|3| can be part of the constant.

Anyway, hope you were not too unpleasantly diverted by the pics.

__________________________________________
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• Feb 5th 2010, 12:48 PM
driver327
I am sorry, I do not understand the graph, this is for Calc II class
• Feb 5th 2010, 12:59 PM
tom@ballooncalculus
Pardon the diversion, anyway I edited in the relevant point for no. 1 above.

2). You aren't clear whether you plugged in values for x or theta, probably x, and maybe your inversing is muddled - integral should be

sin^-1(x/3)

Edit:

Again, just in case it helps...

• Feb 6th 2010, 01:12 PM
driver327
Oh!! I got it now for #1. Thank you!
• Feb 6th 2010, 01:44 PM
Soroban
Hello, driver327!

Quote:

$2)\;\;\int^{\frac{3}{2}}_0\frac{dx}{\sqrt{9- x^2}}$
This is a standard form

. . There is a formula: . $\int\frac{du}{\sqrt{a^2-u^2}} \;=\;\arcsin\left(\frac{u}{a}\right) + C$

If you want to do it out:

. . $\text{Let: }\:x \,=\,3\sin\theta \quad\Rightarrow\quad dx \,=\,3\cos\theta\,d\theta \quad\Rightarrow\quad \sqrt{9-x^2} \,=\,3\cos\theta$

Substitute: . $\int\frac{3\cos\theta\,d\theta}{3\cos\theta} \;=\;\int d\theta \;=\;\theta + C$

Back-substitute: . $3\sin\theta = x \quad\Rightarrow\quad \sin\theta \:=\:\frac{x}{3} \quad\Rightarrow\quad \theta \:=\:\arcsin\frac{x}{3}$
. . So we have: . $\arcsin\frac{x}{3} \,\bigg]^{\frac{3}{2}}_0$

Evaluate: . $\arcsin\left(\frac{1}{3}\cdot\frac{3}{2}\right) - \arcsin\left(\frac{0}{3}\right) \;\;=\;\;\arcsin\frac{1}{2} - \arcsin 0 \;\;=\;\;\frac{\pi}{6} - 0 \;\;=\;\;\frac{\pi}{6}$