trig substitution

**driver327** · February 5th, 2010, 11:55 AM

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

**tom@ballooncalculus** · February 5th, 2010, 12:26 PM

1) With sqrt(1 + x^2) in the denominator you'll find sinh^-1 an easier sub.

Just in case a picture helps...

... where

... 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).

Spoiler:

Ok, your way is equally good, and to answer your question, actually...

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.

__________________________________________
Don't integrate - balloontegrate!
Balloon Calculus; standard integrals, derivatives and methods
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**driver327** · February 5th, 2010, 12:48 PM

I am sorry, I do not understand the graph, this is for Calc II class

**tom@ballooncalculus** · February 5th, 2010, 12:59 PM

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...

Spoiler:

**driver327** · February 6th, 2010, 01:12 PM

Oh!! I got it now for #1. Thank you!

**Soroban** · February 6th, 2010, 01:44 PM

Hello, driver327!

$2)\;\;\int^{\frac{3}{2}}_0\frac{dx}{\sqrt{9- x^2}}$

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

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