Question

pls give me some integration problems (no drawings involved) to work on :)

Answer 1

Ok.

Answer 2

Watch this one.

Answer 3

integral of (sec^6 x)/(tan^2 x) dx

Answer 4

That question came off of one of my midterms

Answer 5

\[\int\limits_{}^{}e ^{-x^2} dx\]

Answer 6

\[\int\limits_{}^{}\sin(x^2) dx\]

Answer 7

\[\int\limits_{}^{}\sin(x^3) dx\]

Answer 8

And finally, the not so meanie one, \[\int\limits_{}^{}\sec ^3 x dx\]

Answer 9

\[
\huge\int e^{\sin x}\cdot \frac{x\cos^3x-\sin x}{\cos^2x}dx
\]

Answer 10

\[\int\limits_{}^{}\Gamma(x)erf(x)dx\]

Answer 11

That last one right there, I'm not sure even if it is valid

Answer 12

\[\mathsf{\text{Prove }\int\limits_0^{2\pi} f(x) \cos x \ge0,\quad \text{given f''(x)} \ge \text{0 and f(x) is continuous on }{[0,2\pi].}}\]

Answer 13

An interesting (but somewhat hard) is:
Find a real number r and a positive number L such that:\[\LARGE\lim_{x \rightarrow \infty} \frac{r^c \int\limits_{0}^{\pi /2} x^r sinx dx}{\int\limits_{0}^{\pi /2}x^r cosxdx } = L\]

Answer 14

@juanjohnguy

\(\LARGE \frac{\tan^3 x}{3} + 2\tan x - \cot x\)

Answer 15

@bmp , can you do the one's I've given?

Answer 16

Typo: should be a real number c and a positive number L.

Answer 17

And the limit is r -> infinity

Answer 18

@lgbasallote , do sec^3 x, it's a fun integration by parts

Answer 19

lol...how come everytime an asker asks for some problems to work on people give those thingies o.O

Answer 20

Because they are fun :-)

Answer 21

You never specified the difficulty.

Answer 22

juan's problem was the only one possible :P

Answer 23

@bmp , how do you do Integrate[Gamma[x]*Erf[x], x]? Not even mathematica knows

Answer 24

sec^3 x is very possible. it's integration by parts

Answer 25

spoiler: u=sec x

Answer 26

Mine's totally possible. Here's an easier one:
\[
\int \frac{dx}{\sqrt{1+e^x}}
\]

Answer 27

Omg, I win. I made the series expansion have the euler-mascheroni constant!

Answer 28

you almost had it, did you want me to share the answer with you?

Answer 29

The sin(x^2) is not that hard. Take the imaginary part: \[\ Im( \int_{0}^{\infty} e^{-ix^2}dx ) = \frac{1}{2}\sqrt{\frac{\pi}{2}}\]I think.

Answer 30

lol +C? hahaha @juanjohnguy

Answer 31

No, it's fresnal s

Answer 32

try the wolf.

Answer 33

i liked my tutorials better :P

Answer 34

well, stuf AND fresnel S

Answer 35

and, isn't that complex analysis or something, @bmp

Answer 36

\[\mathsf{\text{Evaluate,}\large\int\limits_0^1\left(\sqrt[m]{1-x^n} - \sqrt[n]{1-x^m}\right)\;dx}\]

Answer 37

I think so, I never took an analysis course. I learned the cos(x^2) and sin(x^2) because they are quite cool :-)

Answer 38

\[
\int \sec^3x dx=\int(\sec^2x)(\sec x)dx=\tan x \sec x - \int(\tan x)(\sec x \tan x)dx\\
= \tan x \sec x - \int \sec x (\sec^2x-1)dx\\
= \tan x \sec x - \int \sec^3xdx+\int \sec x dx\\
= \frac{1}{2}[\tan x \sec x + \log(\sec x + \tan x)]
\]

Answer 39

lol not quite, the answer that i got was all the same except the beginning is \[(1/2)\tan ^{3}x\]

Answer 40

this thread is laggy :/

Answer 41

How bout sin(x^3)? why does it have Gamma[] when I mathematica it?

Answer 42

I almost considered doing a joke relating the laggy thread with the recent change of picture by @lgbasallote but disconsider.
@inkyvoyd You got me. I will try to find something :-)

Answer 43

Alrighty :P

Answer 44

I can give you another one that was on my old midterm if you like

Answer 45

sure juan..you're the only one who gives possible questions:P haha

Answer 46

I gave possible questions! hey!

Answer 47

lol

Answer 48

\[\mathsf{\int \frac{\sec^2x}{\left(\sec x +\tan x\right)^{9/2}}dx}\]

Answer 49

Wait no. I only gave *a*possible question

Answer 50

only sin^3 x :P

Answer 51

\[\huge
\int x^2 e^xdx\\
\huge\int x^3 e^{x^2} dx\\
\huge\int x^2\sin x dx
\]
Those are all possible

Answer 52

A interesting one then \[ \int \sqrt{\tan{x}}dx \]

Answer 53

Secant ^3 x :P

Answer 54

I like Integrate[Gamma[x]*Erf[x], x] though

Answer 55

lol50 replies srsly :P

@bmp lana gave me that before already haha couldnt solve it

Answer 56

\[\int\limits_{?}^{?} dx/\sqrt{4x+x^2}\]

Answer 57

@bmp , is that by parts or u-sub? I could try it out

Answer 58

@lgbasallote Did you try u = tan(x)? :-)

Answer 59

the question marks don't mean anything

Answer 60

\[\mathsf{\large\int\limits_0^{\frac{\pi}2}\left(\sqrt{\ tan x} + \sqrt{\cot x}\right)\;dx}\]

Answer 61

Wth is PolyGamma?

Answer 62

I did that once @inkyvoyd . Once. I will never redo it for nothing.

Answer 63

But it's a u-sub, at least how I solved it.

Answer 64

yes @bmp

@juanjohnguy \(\LARGE \frac{1}{2} \ln (\frac{x}{\sqrt{4x + x^2}})\)

Answer 65

You mean the prob you gave?

Answer 66

Or wth is polygamma?

Answer 67

No, the sqrt(tan(x)) one @inkyvoyd . And I think no sane person should work this one out...

Answer 68

No sorry thats not right. Just a hint is that you have to use trig substitution for that one.

Answer 69

Feed it to Mathematica and look at the answer. Or use wolfram and check the show steps.

Answer 70

i did..phooey :/

Answer 71

oh i see phooey i removed the sqrt >.<

Answer 72

lol did you complete the square for the 4x+x^2 part?

Answer 73

Ok, after I use the bathroom, finish my bio unit test, I will try it.

Answer 74

sq rt tan x right? Easy to memorize.

Answer 75

Yeah. But so tricky to solve. :-)

Answer 76

it seems i squared after getting the trig counterparts i forgot it was square rooted..final ansewer

\(\Large \ln (\frac{x + 2 + \sqrt{4x + x^2}}{2})\)

Answer 77

All of it is right according to my answer however the "stuff" under the square root is (x+2)^2 + x^2

Answer 78

Just cause at the beginning i completed the square for the 4x + x^2

Answer 79

ohh phooey...guess i have a lot to learn haha

Answer 80

lol but you def almost got it you where extremely close so maybe not too much

Answer 81

I got the same answer as @lgbasallote :-)

Answer 82

Here is another \[\int\limits_{}^{}(1+sinx)/(1-sinx) dx\]

Answer 83

hm....then i guess the integral as 2 answers?

Answer 84

I mean there are dif ways to do a problem

Answer 85

I did complete the square first. What did you got afterwards?

Answer 86

i cant solve it T_T i got stuck in cos^2x/(1-sinx)^2

Answer 87

Maybe u = tan(x/2)? What I got is quite hairy, but it seems solvable.

Answer 88

after i completed the square i got \[\int\limits_{}^{}dx/\sqrt{(x+2)^2 - 2^2}\] then i made my my x+2= 2 sec\[\theta\] dx=2sec\[\theta \tan \theta\] and the square root of all of the stuff under the square root= 2tan\[\theta\]

Answer 89

and hold on let me check my paper and see where you are at in relation to mine igbas

Answer 90

i too lagged :/ im closing this thread now

Answer 91

alright

Answer 92

perhaps open a new one and it wont be as laggy?