Are You Ready to Challenge Your Integral Solving Skills?

[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 I am completely lost, I just want to see where I am 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 don't mind...

 

[ssd]

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

 

[Hootenanny]

Slightly more difficult one; Find I if

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

 

[Gib Z]

O my jesus...I am completely lost on both of those, but from what I see on the mathematicia integrator, which doesn't show me steps btw..The solution to ssd's is way out of my leauge, and i haven't 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]

O my jesus...I am completely lost on both of those, but from what I see on the mathematicia integrator, which doesn't show me steps btw..The solution to ssd's is way out of my leauge, and i haven't 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

 

[Werg22]

O my jesus...I am completely lost on both of those, but from what I see on the mathematicia integrator, which doesn't show me steps btw..The solution to ssd's is way out of my leauge, and i haven't 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).

 

[d_leet]

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?

 

[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]

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

Did you remember to put your code between [tex]\text{$$}$$ and $$\text{[\//tex]}$$ 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, I am 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 haven't 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 I am not right, because I'd still have trouble with that lol. I don't actually know my level, that's why I am 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:

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

 

[Gib Z]

I don't 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 don't 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 $$\int u^{\frac{1}{2}} \frac {2u}{1+u^4} du$$. 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.

 

[cristo]

I don't 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 don't 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]

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.

 

[Gib Z]

O CRAP! I am so so SO sorry! Mental Blank, yes I am 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 doesn't help me, is letting u^2=tan(x/2) gives me $$\int \tan x du$$..

 

[cristo]

O CRAP! I am so so SO sorry! Mental Blank, yes I am 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.

 

[Gib Z]

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 won't be able to finish that today, its almost 12 here and my eyes are sooo red...o my that looks daunting...

 

[cristo]

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 won't 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.

 

[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.

 

[cristo]

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.

I would suspect that the answer given is correct, however, off the top of my head, I don't know the solution to the given integral. There are, of course, many ways to write the same algebraic expression.

 

[Mute]

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 I am not right, because I'd still have trouble with that lol.

You probably won't end up using that form of sinx, but just in case you didn't just make a typo, I thought I'd point our that you're missing a factor of $1/i$:

$$\sin z = \frac{e^{iz} - e^{-iz}}{2i} = -i \sinh(iz)$$

 

[Hootenanny]

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.

You are of course correct, at the moment I tend to go the long way round to no where, especially in calc. Just what you need when your exams are coming up

 

[Gib Z]

Umm ok I am having numerous troubles with the sqrt tan x one, I am giving up..ill get back to it later when i get better..As to the one Hootenanny gave me, if i try the suggestion let z= tan x/2, i need help expressing (sin [x/2])^2 in terms of z, if someone can do that ill be able to keep on going..

EDIT: umm help with doing the same with cos instead of sin would be ok as well.

 

[NTUENG]

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.

Hmm, i tink trigo. substitution may work. U can let $$u^4 = \cos 2\theta , du = \frac{1}{4}(\cos 2\theta)^{-\frac{3}{4}} (-2 \sin 2\theta)$$. Since $$\cos 2\theta = 2 \cos^2 \theta - 1$$, the bottom of the integral simplifies to $$1 + 2 \cos^2 \theta - 1 = 2 \cos^2 \theta$$. But the top can be quite messy. So, u can take your time to evaluate it.

Hope that helps!

 

[Gib Z]

hmm ill be up hours tonight doing that :D ty NTUENG

 

[ssd]

Umm ok I am having numerous troubles with the sqrt tan x one, I am giving up..ill get back to it later when i get better..As to the one Hootenanny gave me, if i try the suggestion let z= tan x/2, i need help expressing (sin [x/2])^2 in terms of z, if someone can do that ill be able to keep on going..

EDIT: umm help with doing the same with cos instead of sin would be ok as well.

Check reply #17 and #25 of where I gave the hint of sqrt(tanx) problem.

 

[LRJ85]

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!

Yes, partial fractions will definitely work. The original integral is $$\int \sqrt{\tan x} dx$$. After letting tan x = u², u convert it to $$2\int \frac{u^2}{1+u^4} du$$. Complete the square at the bottom to get (u²+1)²-2u² = $$(u^2-u\sqrt{2}+1)(u^2+u\sqrt{2}+1)$$. Your integral becomes $$\int \frac{u^2}{(u^2-u\sqrt{2}+1)(u^2+u\sqrt{2}+1)} du$$. By partial fractions, $$\frac{u^2}{(u^2-u\sqrt{2}+1)(u^2+u\sqrt{2}+1)} = \frac{Au+B}{u^2+u\sqrt{2}+1} + \frac{Cu+D}{u^2-u\sqrt{2}+1}$$. Compare coefficients on both sides to get A=$$-\frac{1}{2\sqrt{2}}$$, C=$$\frac{1}{2\sqrt{2}}$$ and B=D=0. After that, manipulate the integral by factorization, completing the square and expressing the result in terms of ln and $$\tan^{-1}$$, finally u get the answer $$\frac{\sqrt{2}}{8} ln \frac{u^2-u\sqrt{2}+1}{u^2+u\sqrt{2}+1} + \frac{\sqrt{2}}{4} \tan^{-1} (u\sqrt{2}-1) - \tan^{-1} (u\sqrt{2}+1) + C$$. The last step is to convert u back to original variable x using the relation u² = tan x and u get your answer in terms of x.

Haha, isn't that a good solution ??!

 

[Gib Z]

Thats really good, thanks for that. For the partial fractions one, I actually managed to get A and C before, but I could only reduce it to B+D=0, I couldn't acutally get the values themselves. You could explain how you got that one?

Even if I did get B and D, What steps did you do to get to the next result? I have no idea...sorry about that.

 

[Gib Z]

As for ssd's let's z= sin x - cos x...i can't see how to manipulate your expression for that to help me..