Past exam questions

**obesechicken13** · May 15th 2010, 09:41 AM

So I just passed my first year, but I failed calc 2... badly
I'll have to retake it next summer, but until then I would like to ask some questions that I saw on the 2009 exam.

The past exams can all be found at this website
Skule Courses - SKULE
the course is mat197

If you would like to, you can read it there for more clarity.

1. Evaluate the integral:
erhm... could anyone send me a link of the math functions?

Anyways, the integral of
x^2 dx / (x^2+64)^(3/2)

I first thought, maybe I should try some substitution, but that didn't work. The bottom looked eerily familiar to an inverse trig substitution style, that didn't work either. Then I thought, maybe I need to pull something out, like that 64=2^6 and (2^6)^3/2 = 2^9... and we could use that to some strange effect like completing the square.

The entire exam is attached to this post. Although you are welcome to do every single problem, it is not required, as I would like to tackle them a little more in depth than I had previously, before posting them. Many thanks to anyone who answers, to help this student in need.

**AllanCuz** · May 15th 2010, 10:24 AM

Originally Posted by obesechicken13

So I just passed my first year, but I failed calc 2... badly
I'll have to retake it next summer, but until then I would like to ask some questions that I saw on the 2009 exam.

The past exams can all be found at this website
Skule Courses - SKULE
the course is mat197

If you would like to, you can read it there for more clarity.

1. Evaluate the integral:
erhm... could anyone send me a link of the math functions?

Anyways, the integral of
x^2 dx / (x^2+64)^(3/2)

I first thought, maybe I should try some substitution, but that didn't work. The bottom looked eerily familiar to an inverse trig substitution style, that didn't work either. Then I thought, maybe I need to pull something out, like that 64=2^6 and (2^6)^3/2 = 2^9... and we could use that to some strange effect like completing the square.

The entire exam is attached to this post. Although you are welcome to do every single problem, it is not required, as I would like to tackle them a little more in depth than I had previously, before posting them. Many thanks to anyone who answers, to help this student in need.

$\int \frac{ x^2 } { (x^2+64)^{ \frac{3}{2} } } dx = \int \frac{ x^2 } { (x^2+8^2)^{ \frac{3}{2} } } dx$

Let $x = 8tan f$ and $dx = 8sec^2 f df$

$\int \frac{ x^2 } { (x^2+8^2)^{ \frac{3}{2} } } dx$

$\int \frac{ 64 tan^2 f (8sec^2 f) }{ ( 8^2[tan^2 f + 1] )^{ \frac{3}{2} } } df$

$\int \frac{ 64 tan^2 f (8sec^2 f) }{ ( 8^2 sec^2 f )^{ \frac{3}{2} } } df$

$\int \frac{ 64tan^2 f (8 sec^2 f) }{ ( 8^ 3sec^3 f ) } df$

$\int \frac{ tan^2 f }{ sec f } df$

$\int \frac{ tan^2 f }{ \frac{1}{cosf} } df$

$\int \frac{ \frac{sin^2 f}{cos^2 f} }{ \frac{1}{cosf} } df$

$\int \frac{sin^2f}{cosf} df$

Can you solve from here?

**Soroban** · May 15th 2010, 10:33 AM

Hello, obesechicken13!

$[1]\;\;\int \frac{x^2}{(x^2+64)^{\frac{3}{2}}}\,dx$

Let: . $x \:=\:8\tan\theta \quad\Rightarrow\quad dx \:=\:8\sec^2\!\theta\,d\theta$

Substitute: . $\int\frac{(8\tan\theta)^2}{(64\sec^2\!\theta)^{\fr ac{3}{2}}}(8\sec^2\!\theta\,d\theta) \;=\;\int\frac{512\tan^2\!\theta\sec^2\!\theta}{51 2\sec^3\!\theta}\,d\theta$

. . . $=\;\;\int\frac{\tan^2\!\theta}{\sec\theta}\,d\thet a \;\;=\;\;\int\frac{\sec^2\!\theta-1}{\sec\theta}\,d\theta$

. . . $=\;\;\int\left(\sec\theta - \cos\theta\right)\,d\theta \;\;=\;\;\ln|\sec\theta + \tan\theta| - \sin\theta + C$

Back-substitute: . $\tan\theta \:=\:\frac{x}{8} \quad\Rightarrow\quad \sec\theta \:=\:\frac{\sqrt{x^2+64}}{8}$

We have: . $\ln\left|\frac{\sqrt{x^2+64}}{8} + \frac{x}{8}\right| - \frac{x}{\sqrt{x^2+64}} + C \;\;=\;\;\boxed{\ln\left|x + \sqrt{x^2+64}\right| + \frac{x}{\sqrt{x^2+64}} + C}$

**obesechicken13** · May 15th 2010, 07:00 PM

Firstly, thanks to both of you.

Can you solve from here?

Yes, I can by substituting 1-cos^2(x) in for sin^2(x) and then splitting the integral. Then I draw a triangle to back substitute for x.
I arrive at the same answer as the second to last step as Soroban

Originally Posted by Soroban

We have: . $\ln\left|\frac{\sqrt{x^2+64}}{8} + \frac{x}{8}\right| - \frac{x}{\sqrt{x^2+64}} + C \;\;=\;\;\boxed{\ln\left|x + \sqrt{x^2+64}\right| + \frac{x}{\sqrt{x^2+64}} + C}$

I don't understand how you went from the second to last step to the last step. How did you get rid of the 8's and how did the subtraction turn into addition?

PS: I no longer wish to learn the math code, it seems not worth the trouble
In addition, I would like to know exactly how I should go about an integration problem, but I suppose that will come with practice. So never mind about that.

One thing this problem has taught me is how little I know my trigonometric identities and their integrals. I'm gonna make some flashcards.

**obesechicken13** · May 16th 2010, 09:12 AM

Perhaps the second to last step is a good enough solution that it doesn't need to go the last step. Thanks!

**Chris L T521** · May 16th 2010, 09:23 AM

Originally Posted by obesechicken13

Perhaps the second to last step is a good enough solution that it doesn't need to go the last step. Thanks!

Observe that

$\begin{aligned}\ln\left|\frac{\sqrt{x^2+64}}{8} + \frac{x}{8}\right| - \frac{x}{\sqrt{x^2+64}} + C &= \ln\left|\frac{x+\sqrt{x^2+64}}{8}\right|-\frac{x}{\sqrt{x^2+64}}\\ &= \ln\left|x+\sqrt{x^2+64}\right|-\ln 8-\frac{x}{\sqrt{x^2+64}}+C\end{aligned}$

But $C-\ln 8$ is just another constant!! We can rename it $C$ and then get the simplified result Soroban got (except I just noticed a small typo on his part.. :/ ).

Does this make sense?

**obesechicken13** · May 16th 2010, 09:57 AM

Yes it does, that typo threw me off

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