I fucked up my first cal 2 test of the semester and I want to get better for my test next week.
ITT: The hardest cal 2 questions you know / cal 2 questions you've fucked up on
pic related, the two questions that fucked me over last test
>>7950511
[math] \displaystyle \int{ \sqrt { \tan{ x} } \mathbb{d}x } [/math]
[math] \displaystyle \int{ \frac {1} { \sin{x} } \mathbb{d}x } [/math]
Knock yourself out.
>>7950608
t = tan(x/2) solves everything
>>7950608
1/sin(x) = sin(x)/sin(x)^2 = sin(x)/(1-cos(x)^2)
Then U sub
replace the tan(x) with sin(x)/cos(x) then u sub
2ezm8
>>7950622
I prefer multiplying it by cscx+cotx/cscx+cotx and doing it that way
>>7950608
Still stuck on the root of tanx problem though, that's a good one
If [math]a[/math] and [math]b[/math] are positive numbers, what is
[eqn] \int_0^\infty \frac{ e^{ax} - e^{bx} }{ (1+ e^{ax}) (1+ e^{bx}) } \; \mathrm{d}x [/eqn]
>>7950511
http://www.math.lsa.umich.edu/courses/116/Exams/index.html
Have fun
>>7950658
>http://www.math.lsa.umich.edu/courses/116/Exams/index.html
wew lad, not even a single theoretical question on the first exam. just dumb calculations, and they even let you use a ti-89. too easy.
[math]\frac{1}{\sqrt{2\pi}} \int_{-\infty }^{\infty} e^{-\frac{1}{2}x^2} dx[/math]
You should be able to solve this.
>>7950511
Do you have to prove the integral of secx or are you allowed to memorize it? Because that first integral is fucking obnoxious if you can't.
looking at the previous exams from fall 2015 looks like youre fucked for the next exam/final unless you can figure out a way to bring your phone into the room during the test.
As fun as solving integrals is, it's a shame it's an outdated waste of time.
>>7950654
[math]\int_0^\infty \frac{ e^{ax} - e^{bx} }{ (1+ e^{ax}) (1+ e^{bx}) } \; \mathrm{d}x[/math]
[math]\int_0^\infty \frac{ e^{ax} - e^{bx} + 1 - 1}{ (1+ e^{ax}) (1+ e^{bx}) } \; \mathrm{d}x[/math]
[math]\int_0^\infty \frac{ (1 + e^{ax}) - (1 + e^{bx})}{ (1+ e^{ax}) (1+ e^{bx}) } \; \mathrm{d}x[/math]
[math]\int_0^\infty \frac{1}{1+e^{bx}} \; \mathrm{d}x - \int_0^\infty \frac{1}{1+e^{ax}} \; \mathrm{d}x[/math]
Now just solve
[math]\int_0^\infty \frac{1}{1+e^{ax}} \; \mathrm{d}x[/math]
Let [math]u = 1 + e^{ax}[/math], [math]\mathrm{d}u = ae^{ax} \mathrm{d}x[/math]
[math]e^{ax} = u - 1[/math], so [math]\mathrm{d}x = \frac{\mathrm{d}u}{a(u - 1)}[/math]
So, [math]\int_0^\infty \frac{1}{1+e^{ax}} \; \mathrm{d}x = \lim_{n\to\infty} \int_0^n \frac{1}{1+e^{ax}} \; \mathrm{d}x = \frac{1}{a}\lim_{n\to\infty} \int_2^{1+e^{an}} \frac{1}{u^2-u} \; \mathrm{d}u[/math]
Split using partial fractions:
[math]\frac{1}{a}\lim_{n\to\infty} \int_2^{1+e^{an}} \frac{1}{u^2-u} \; \mathrm{d}u = \frac{1}{a}\lim_{n\to\infty} (\int_2^{1+e^{an}}\frac{1}{u-1}\;\mathrm{d}u - \int_2^{1+e^{an}}\frac{1}{u}\;\mathrm{d}u)[/math]
[math]=\frac{1}{a}\lim_{n\to\infty}((\ln(e^{an})-\ln(1)) - (\ln(1+e^{an})-\ln(2)))[/math]
[math]=\frac{1}{a}\lim_{n\to\infty}(\ln(e^{an})- \ln(1+e^{an})+\ln(2)) = \frac{\ln(2)}{a}[/math]
So, [math]\int_0^\infty \frac{1}{1+e^{bx}} \; \mathrm{d}x - \int_0^\infty \frac{1}{1+e^{ax}} \; \mathrm{d}x = \ln(2)(\frac{1}{b} - \frac{1}{a})[/math]
Whew. My calc is a bit rusty so let me know if I got anything wrong.
>>7950511
I already finished my calc 2 finals but I still have access to my professor's practice tests and solutions if you want them.
>>7950744
Bruh. Anti derivative of sec x is easy as fuck. Dxsec = secxtanx and Dxtanx = secx*secx. Blah blah blah FTC implies
Anti derivative of secx = ln (secx + tanx) + c
>>7950765
(You)
>>7950744
Mux. and divide by [math]\sec x + \tan x[/math]... (essentially just numerator and denominator multiplication)
[eqn]
\begin{aligned}
\int \sec x \; dx &= \int \sec x \frac{\sec x + \tan x}{\sec x + \tan x} \; dx \\
u = \sec x + &\tan x \quad du = \sec x + \tan x + \sec^2 x \; dx \\
&= \int \frac{\sec x + \tan x + \sec^2 x \; \; dx}{\sec x + \tan x} \\
&= \int \frac{du}{u} \\
&= \ln |u| + C \\
&= \ln | \sec x + \tan x | + C
\end{aligned}
[/eqn]
The same applies to [math]\int \csc x \; dx[/math]... Pay close attention, the operations are inverted, too.
[eqn]
\begin{aligned}
\int \csc x \; dx &= \int \csc x \frac{\csc x + \cot x}{\csc x + \cot x} \; dx \\
u = \csc x + &\cot x \quad du = - \csc x \cot x - \csc^2 x \; dx \\
&= \int \frac{- \csc x \cot x - \csc^2 x \; \; dx}{\csc x + \cot x} \\
&= \int - \frac{du}{u} \\
&= - \ln |u| + C \\
&= - \ln | \csc x + \cot x | + C
\end{aligned}
[/eqn]
>>7950765
Did you really figure that out or what?
>>7950885
>Did you really figure that out
Yes.
>>7950765
can you separate multiples like that? when it's (1+e^(ax))(1+e^(bx)) as a dividend
>>7951036
Yes
>>7951036
Basically, what he did was noted that
[eqn]
\frac{b-a}{ab} = \frac{1}{a} - \frac{1}{b}.
[/eqn] It's the same thing if you replace the variables with the exponents here.
>>7951036
This is how partial fraction decomposition works.
[eqn]\frac{1}{(x+a)(x+b)}
= \frac{1}{a-b} \frac{a-b}{(x+a)(x+b)}
= \frac{1}{a-b} \frac{(x+a)-(x+b)}{(x+a)(x+b)}
= \frac{1}{a-b} \left( \frac{1}{x+b} - \frac{1}{x+a} \right)
[/eqn]
For the second part of that integral you can do (for a>=0):
[eqn]\int_0^\infty \frac{1}{1+e^{a x}} dx
= \frac{1}{a} \int_0^\infty \frac{1}{1+e^x} dx
= \frac{1}{a} \int_0^1 \frac{1}{1+\frac{1}{x}} \frac{1}{x} dx
= \frac{1}{a} \int_0^1 \frac{1}{x+1} dx
= \frac{1}{a} log(2)
[/eqn]
>>7951089
Whoops should be a>0 since otherwise division by 0.
>>7950765
>my calc rusty
>not chugging straight into W|A
>>7950635
This, but some would say that's just memorization
>>7950511
calculus is hard
>>7950608
The second one is literally just integrating csc x
can any of you anons do the first question in the OP. That is literally going to be on my test
>>7951701
Ask in /sqt/, they'll help you.
>>7950608
There's probably an easier way to solve this.
[eqn]
\begin{align}
\int^x \sqrt{ \tan(t) } dt
&= \int^{\tan(x)} \frac{ \sqrt{t} }{ t^2+1 } dt \\
&= \int^{ \sqrt{\tan(x)} } \frac{2 t^2}{ t^4+1 } dt \\
&= \int^{ \tan^{-1}( \sqrt{\tan(x)} ) } \frac{2 \tan(t)^2 \sec(t)^2 }{ \tan(t)^4+1 } dt \\
&= 2 \int^{ \tan^{-1}( \sqrt{\tan(x)} ) } \frac{1-\cos(2t)}{ 1+\cos(2t)^2 } dt \\
&= \int^{ 2 \tan^{-1}( \sqrt{\tan(x)} ) } \frac{1-\cos(t)}{ 1+\cos(t)^2 } dt \\
&= \int^{ 2 \tan^{-1}( \sqrt{\tan(x)} ) } \frac{1}{ 1+\cos(t)^2 } dt - \int^{ 2 \tan^{-1}( \sqrt{\tan(x)} ) } \frac{\cos(t)}{ 1+\cos(t)^2 } dt \\
&= \int^{ 2 \tan^{-1}( \sqrt{\tan(x)} ) } \frac{\sec(t)^2}{ \tan(t)^2 + 2 } dt - \int^{ 2 \tan^{-1}( \sqrt{\tan(x)} ) } \frac{\cos(t) dt}{ 2-\sin(t)^2 } \\
&= \int^{ \tan( 2 \tan^{-1}( \sqrt{\tan(x)} ) ) } \frac{1}{ t^2 + 2 } dt - \int^{ \sin( 2 \tan^{-1}( \sqrt{\tan(x)} ) ) } \frac{1}{ 2-t^2 } dt \\
&= \frac{1}{\sqrt{2}} \left(
\tan^{-1} \left( \frac{ \tan(2 \tan^{-1}( \sqrt{\tan(x)} ) ) }{\sqrt{2}} \right) -
\tanh^{-1} \left( \frac{ \sin(2 \tan^{-1}( \sqrt{\tan(x)} ) ) }{\sqrt{2}} \right)
\right)
\end{align}
[/eqn]
>>7951701
For sec(x)^3 integral, integrate by parts to reduce to sec(x) integral.
For op's second integral it's just standard complete the square then trig sub.
>>7951395
https://www.mediafire.com/folder/hzqahn85cqrnc/Calc
>>7951701
I didn't include the integral of secx because some anons ITT have already done it.
>>7952226
Integration is trippy
>>7952981
Really complicated improper integrals always fuck my shit up, especially when you throw in l'hôpital's rule and other shit into the mix
>>7952981
Trig integrals usually involve some silly tricks if they're not just sin and cos.
>>7952995
This so much.
>>7952998
I figured, considering the fuckery that's [math]\int \csc x \; dx[/math] and [math]\int \sec x \; dx[/math]. Idk about you and the other anon but solving them with partial fractions makes the process far more understandable (and trackable) as opposed to multiplying and dividing by [math]\csc x + \cot x[/math] and [math]\sec x + \tan x[/math]. It is obviously longer but at least it would be far easier to explain to other people.
>>7950695
>posting meme integral
>>7950695
Standard Gaussian integral, so easy lmao
>>7953233
All integrals are meme integrals.
>>7953984
t. Riemann
>>7950511
The first one is pretty easy. It's just standard IBP
>>7953984
true dat
grad EE here i dont even know how to begin OP's problems
>>7954245
How do we defeat the engineer menace?
>>7950511
perhaps not hard, but definitely something I fucked up on
Decide if the following integral is convergent
[math]
\int_0^\infty( \frac{1}{x} - sin \frac{1}{x}) arctan \space x \space dx
[/math]
>>7954284
Meh, using a proper tex format and another integral that is calc 1 tier but nice to revise substitution.
[math]
\begin{aligned} \int_0^\infty (\frac{1}{x} - \sin{\frac{1}{x}}) \arctan{x}\; dx \end{aligned}
\\
\begin{aligned} \int \frac{1}{(t^2+1)^n} \; dt \end{aligned}
[/math]
MIT integration bee
you should be able to solve this
>>7954317
K-kurisu?
>>7954300
Anybody interested in knowing the solution?
[math] \displaystyle \int{ \sqrt { \frac { x^2 + 1} {(x+1)(x+2)} } \mathbb{d}x } [/math]
Also solve the following differential eq:
y=2xy'+1/2*(y')^2-3y'+2
>>7954284
It converges to [math](\gamma+\Gamma(0,1)-\frac{1}{e}) \frac{\pi}{2}[/math].
>>7950511
I'm from /int/ and what the fuck is this shit? :DDD
Seriously though, business law here. Glad I'm done with the mathematical stuff already.
>>7954300
[eqn]\begin{align}\int_0^\infty \left(\frac{1}{x} - \sin \left( \tfrac{1}{x} \right) \right)\arctan x \, \mathrm{d}x &= \int_0^1 \left(\frac{1}{x} - \sin \left( \tfrac{1}{x} \right) \right)\arctan x \, \mathrm{d}x + \int_1^\infty \left(\frac{1}{x} - \sin \left( \tfrac{1}{x} \right) \right)\arctan x \, \mathrm{d}x \\ &= \int_1^\infty \color{red}{\frac{x-\sin x}{x^2}} \, \color{blue}{\arctan \left(\tfrac{1}{x}\right)} \, \mathrm{d}x + \int_1^\infty \color{green}{\left(\frac{1}{x} - \sin \left( \tfrac{1}{x} \right) \right)}\color{orange}{\arctan x} \, \mathrm{d}x \\ &< \int_1^\infty \color{red}{\frac{x+1}{x^2 }}\color{blue}{\frac{1}{x}} + \color{green}{\frac{1}{6x^3}} \color{orange}{\frac{\pi}{2}}\, \mathrm{d}x = \frac{3}{2} + \frac{\pi}{24} \end{align}[/eqn]Assuming n > 1, [eqn]\begin{align} \int \frac{\mathrm{d}t}{(t^2+1)^n} &= \int \frac{\sec^2 u \, \mathrm{d}u}{(\sec^2 u)^n} \\ &= \int \cos^{2n-2}u \, \mathrm{d}u \\ &= \frac{\sin u \cos^{2n-3} u}{2n-2} + \frac{2n-3}{2n-2} \int \cos^{2n-4} u \, \mathrm{d}u \\ &= \frac{1}{2n-2} \frac{t}{(t^2 + 1)^{n-1}} + \frac{2n-3}{2n-2}\int \frac{\mathrm{d}t}{(t^2 + 1)^{n-1}} \end{align}[/eqn]
