# How Good Am I?

1. Jan 1, 2007

### 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...

2. Jan 1, 2007

### ssd

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

Last edited: Jan 1, 2007
3. Jan 1, 2007

### Hootenanny

Staff Emeritus
Slightly more difficult one; Find I if

$$I = \int \frac{dx}{2+\sin(x)}$$

4. Jan 1, 2007

### 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?

5. Jan 1, 2007

### cristo

Staff Emeritus
Why not try the substitution u2=tanx ?

Hint: Remember that sec2x=1+tan2x

Last edited: Jan 1, 2007
6. Jan 2, 2007

### Werg22

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

7. Jan 2, 2007

### d_leet

How exactly do you plan on doing that?

8. Jan 2, 2007

### Werg22

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

9. Jan 2, 2007

### Von Gastl

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

10. Jan 2, 2007

### Swapnil

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

Last edited: Jan 2, 2007
11. Jan 2, 2007

### Pathway

12. Jan 2, 2007

### Hootenanny

Staff Emeritus
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.

13. Jan 2, 2007

### 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 $$\int sqrt{u} \cdot \frac {1}{1+u^2} du$$, 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 $$\frac {2x^{3/2}}{3(x^2+1)} + \frac {8x^{7/2}}{21}$$ 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 $$\sin x = \frac {e^{ix} - e^{-ix}}{2}$$ which I got from Eulers Identity, looks like $$\sinh ix$$ 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..

Last edited: Jan 2, 2007
14. Jan 2, 2007

### cristo

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

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

15. Jan 2, 2007

### Gib Z

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

Last edited: Jan 2, 2007
16. Jan 2, 2007

### Gib Z

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

17. Jan 2, 2007

### 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.

Last edited: Jan 2, 2007
18. Jan 2, 2007

### cristo

Staff Emeritus
I'm not pissed; sorry if my post seemed a bit blunt! HAve you come across the chain rule for differentiation? It states that $$\frac{d}{dx}(f)=\frac{df}{du}\frac{du}{dx}.$$

So, $$\frac{d}{dx}u^2=\sec^2x \Rightarrow \frac{d(u^2)}{du}{\frac{du}{dx}=\sec^2x \Rightarrow 2u \frac{du}{dx}=\sec^2x \Rightarrow 2u du=\sec^2x dx$$

19. Jan 2, 2007

### cristo

Staff Emeritus
I think you have one mistake here, namely the $\sqrt u$. Since $u^2=tanx$ this should simply be u. I don't think you need any more substitution or int. by parts- just use partial fractions.

20. Jan 2, 2007

### Gib Z

O CRAP! I am so so SO sorry! Mental Blank, yes im completely fine with differentiation. O thank you. Ok Well that ends up giving me $$2\int \frac {u^2}{1+u^4} du$$ Umm...I don't get how you would do it with partial fractions, i was thinking of trig sub, i duno how though. What i did notice though, even though it probably doesnt help me, is letting u^2=tan(x/2) gives me $$\int \tan x du$$..

Last edited: Jan 2, 2007