How Good Am I?

Gib Z

Hey guys I was just wondering If you guys could post over a few indefinite integrals for me to solve, because I seem to find some of the integrals on these forums easy, and others im completely lost, I just want to see where im up to so I can start learning from there, my learning of calculus has been independant, so it hasnt been structured very well...

Just post maybe a few at a time, with varying difficulty and I'll try to do them. I would prefer to get advice and do them on this thread rather than learn from a link if you guys dont mind...

ssd

Start with one of 12th standard level (it was in our text): sqrt(tanx)

Hootenanny

Slightly more difficult one; Find I if

[tex]I = \int \frac{dx}{2+\sin(x)}[/tex]

Gib Z

O my jesus...I am completely lost on both of those, but from what I see on the mathematicia integrator, which doesnt show me steps btw..The solution to ssd's is way out of my leauge, and i havent even bothered to look up Hootenannys if its harder..Could you gives give me a prod in the right direction? Usually when I answer peoples integrals on this site I've read the hints others have given before...Im thinking maybe t=tan(x/2) substitution?

cristo

Why not try the substitution u²=tanx ?

Hint: Remember that sec²x=1+tan²x

Werg22

I would start with a change of bounds to simplify the integral. That is to say, find the integral of cosec(x).

d_leet

How exactly do you plan on doing that?

Werg22

Dah! Forget it. For some reason I thought it was 1/sin(x+2). My bad.

Von Gastl

Here something close:
intergrat (1/(1+sin(u)) du = tan u - sec u +C

I'm having trouble with LaTeX code reference loading up.

Swapnil

Did you remember to put your code between [tex]\text{[tex]}[/tex] and [tex]\text{[\//tex]}[/tex] tags?

Pathway

This may help.

http://www.math.wisc.edu/~nagel/Math275Assignments.html

Part of 4, all of 5-7. Textbook exercises are from Tom M. Apostol's Calculus, volume 1. Not all of them are from the text, though.

Hootenanny

My apologies Gib if my integral seemed a little overwhelming, it's just one of the standard substitutions that I seem to remember. Perhaps if you told us your level we could set questions more appropriately.

P.S. For my integral, a hyperbolic substitution would do the trick.

Gib Z

Umm, Im not sure If i got it, But basically for ssd's one I let u= tan x, because his u^2=tan x didnt seem to work for me..Anyway I got it to [tex]\int sqrt{u} \cdot \frac {1}{1+u^2} du[/tex], which I applied integration by parts to. I didn't want to get confused between my u substitution and my u in by parts, so i changed the u to an x for my working in parts and I basically got [tex]\frac {2x^{3/2}}{3(x^2+1)} + \frac {8x^{7/2}}{21}[/tex] where x = tan x, I know i probably havent chosen x again...might confuse some people...but you get the point. And I've never done a hyperbolic substitution before, unless you mean [tex]\sin x = \frac {e^{ix} - e^{-ix}}{2}[/tex] which I got from Eulers Identity, looks like [tex]\sinh ix[/tex] to me. Please tell me im not right, because I'd still have trouble with that lol. I dont actually know my level, thats why im doing this thread. Thanks for all your help guys.

EDIT: I just tried my result for ssd's one in mathematica, didnt turn out nicely..

cristo

Like I said, try the substitution u²=tanx. Why didn't this work for you? Did you substitute for dx correctly? Note:

[tex] 2u du=sec^2(x)dx \Rightarrow 2u du = (1+tan^2(x))dx \Rightarrow dx=\frac{2u}{1+u^4} du [/tex] Substitute this in and evaluate the integral.

Gib Z

I dont seem to understand how you got the first part in your Note. [tex] u^2 = \tan x[/tex] Then [tex]\frac {d}{dx} u^2 = \sec^2 x[/tex] then [tex] d(u^2) = \sec^2 x dx[/tex]. I dont know how you got 2udu...

Sorry for not being good at this guys, I can tell Cristo's a little pissed lol, sorry...

Gib Z

Well from what you showed me, I would get [tex]\int u^{\frac{1}{2}} \frac {2u}{1+u^4} du[/tex]. Just abit of Intergration by parts and trig substitution would get me home free, yea?

ssd

Hint:
sqrt(tanx)= (1/2)[{sqrt(tanx) + sqrt(cotx)} + {sqrt(tanx) - sqrt(cotx)}]
Consider the first part:
sqrt(tanx) + sqrt(cotx)
To integrate this put z= sinx-cosx , then use, 2sinx.cosx= 1-(sinx-cosx)^2.
The second part follows in a similar manner.
I have given the result of the first part in reply #26 of this thread.