# How Good Am I at Integrals?

Homework Helper
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)

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Hootenanny
Staff Emeritus
Gold Member
Slightly more difficult one; Find I if

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

Homework Helper
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
Staff Emeritus
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?
Why not try the substitution u2=tanx ?

Hint: Remember that sec2x=1+tan2x

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

I would start with a change of bounds to simplify the integral. That is to say, find the integral of cosec(x).
How exactly do you plan on doing that?

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

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

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

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Hootenanny
Staff Emeritus
Gold Member
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.

Homework Helper
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..

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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.

Homework Helper
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...

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Homework Helper
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?

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.

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cristo
Staff Emeritus
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...
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, $$\frac{d}{dx}u^2=\sec^2x \Rightarrow \frac{d(u^2)}{du}{\frac{du}{dx}=\sec^2x \Rightarrow 2u \frac{du}{dx}=\sec^2x \Rightarrow 2u du=\sec^2x dx$$

cristo
Staff Emeritus
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?
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.

Homework Helper
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 $$\int \tan x du$$..

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cristo
Staff Emeritus
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.
I don't think a trig sub will work. Try writing the denominator as $(u^2-\sqrt 2u+1)(u^2+\sqrt 2u+1)$, then use partial fractions. It should work, but it won't be pretty!

Edit: In response to your edit, I can't see how it will help, as you have a function of x, integrated wrt u.

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Homework Helper
O God that thought passed through my head, but I thought the answer was going to be nice so I assumed I was wrong lol...my god I wont be able to finish that today, its almost 12 here and my eyes are sooo red...o my that looks daunting...

cristo
Staff Emeritus
O God that thought passed through my head, but I thought the answer was going to be nice so I assumed I was wrong lol...my god I wont be able to finish that today, its almost 12 here and my eyes are sooo red...o my that looks daunting...

Yea, it's not going to be nice. If you put it into the integrator http://integrals.wolfram.com/index.jsp you can see that the solution is not at all simple. Quite a harsh question, if you ask me!!

ssd
Slightly more difficult one; Find I if

$$I = \int \frac{dx}{2+\sin(x)}$$
I feel this is simpler than sqrt(tanx). Putting tan(x/2) = z and writing

sin(x) = 2tan(x/2)/[1+{tan(x/2)}^2]

the result easily follows.

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ssd
Yea, it's not going to be nice. If you put it into the integrator http://integrals.wolfram.com/index.jsp you can see that the solution is not at all simple. Quite a harsh question, if you ask me!!
Does the mentioned link always give correct answer? Because the integral of sqrt(tanx)+sqrt(cotx) is sqrt(2).sin_1(sinx-cosx)+c, by "sin_1(x)" I mean "sine-inverse x" and c is the constant of integration. This answer does not look as frightening as the answer given by the link.

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