Are You Ready to Challenge Your Integral Solving Skills?

[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..

 

[LRJ85]

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.

Ok when u multiply the partial fraction identity by their common denominator, u get $$u^2 = (A+C)u^3 + (B+D+(C-A)\sqrt{2})u^2 + (A+C+(D-B)\sqrt{2})u + B + D$$. Compare coefficients of descending powers of u, u get A+C =0, B+D+(C-A)$$\sqrt{2}$$=1, A+C+(D-B)$$\sqrt{2}$$=0 and B+D=0. Sub. eqn 1 into eqn 3, u get (D-B)$$\sqrt{2}$$=0. Thus D-B must be 0. B=D=0. (shown!)
For the next part, u use the values of A,B,C and D to express the integral as partial fractions. The resultant integral is $$\int \frac{-\frac{u}{2\sqrt{2}}}{u^2+u\sqrt{2}+1} + \frac{\frac{u}{2\sqrt{2}}}{u^2-u\sqrt{2}+1} du$$. U can factor out the constant $$\frac{1}{2\sqrt{2}}$$. After that, try to make the top a derivative of the bottom into the form f'(u)/f(u) where the result is in terms of ln function. For the denominator, u need to complete the square and u get integral of the form $$\int \frac{du}{u^2+a^2}$$ where the result is in terms of $$\tan^{-1}$$ function. Rationalise all the terms with surds at the bottom and use identity ln a - ln b= ln(a/b) to combine the 2 ln terms together. U'll eventually get the answer in thread #33.

 

[ssd]

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..

Write sqrt(tanx) = sqrt(sinx)/sqrt(cosx) and similarly write cotx. Simplify sqrt(tanx)+sqrt(cotx)... check that the numerator = dz/dx=(sinx+cosx). Write the denominator as mentioned in reply #17.

 

[Gib Z]

Hmm ok I've managed to do it the u^2=tan x method, thanks guys. But i want to learn ssd's method, it looks simpler and shorter, and seeing as he posted the question I think he might know how to do it the best, no offense guys. ssd: When I make the numberator of the first part dz/dx, i get the 2nd parts numerator being z, I can't work out how to finish this off.

 

[ssd]

Hmm ok I've managed to do it the u^2=tan x method, thanks guys. But i want to learn ssd's method, it looks simpler and shorter, and seeing as he posted the question I think he might know how to do it the best, no offense guys. ssd: When I make the numberator of the first part dz/dx, i get the 2nd parts numerator being z, I can't work out how to finish this off.

Do the two parts separately.

Put u= sinx+cosx in the second part. You get -du/dx=sinx-cosx= the numerator of the second part. Use 2sinx.cosx = (sinx+cosx)^2 -1.
You appear right in the sense that I feel its the simplest and shortest method when the answer is not known beforehand.

 

[Gib Z]

Hey, I know its been a while since I've posted on this thread, sorry about that guys. Ok well, Could someone give me another integral? Maybe one abit easier than that sqrt tanx

 

[Hootenanny]

No worries. Okay a simpler one for you; Evaluate the following integral

$$\int\frac{8\; dx}{\left(4x^2+1\right)^2}$$

 

[Gib Z]

I used some trig substitution and got $$\int (\frac{2}{\sec \theta})^2 d\theta$$. Lost from there..

 

[Hootenanny]

Which substitution did you use? I would recommend the following one;

$$x:=\frac{1}{2}\tan\theta$$

 

[Gib Z]

$$\int \frac{8dx}{\sec^2 \theta}$$...still lost

 

[Gib Z]

O wait, I get $$\int 4 d\theta$$...I don't Understand what to do now..the answers 4theta? What..

 

[Hootenanny]

O wait, I get $$\int 4 d\theta$$

Correct!

...I don't Understand what to do now..the answers 4theta? What..

You now have to do the integration with respect to theta. What happens when you integrate a number with respect to something?

 

[dextercioby]

Yes, and now express theta in terms of the initial variable, x. Add the integration constant.

Daniel.

 

[Gib Z]

So is the solution $$4\arctan 2x$$...Hey wait, i derived it, it works :), anuda one?

Edit: Forgot the +C

 

[Hootenanny]

Looks good to me.

 

[Hootenanny]

And now for something a little different;

$$I = \int x^2\sin x\;dx$$

 

[Gib Z]

Awesome :) Let's try another :)

Edit: didnt see that post

 

[murshid_islam]

I used some trig substitution and got $$\int (\frac{2}{\sec \theta})^2 d\theta$$. Lost from there..

you were correct. it comes out to be $$\int \frac{4}{\sec^{2} \theta} d\theta$$ which is equal to $$\int 4\cos^{2} \theta d\theta$$. now use the identity $$2\cos^{2}\theta = 1 + \cos(2\theta)$$

 

[Gib Z]

I did integration by parts a few times, I got $$I=\frac{\sin^2 x}{2} - x^2 \cos x$$ :D I think that's right.

Note: This ones for x^2 sin x dx.

 

[Gib Z]

you were correct. it comes out to be $$\int \frac{4}{\sec^{2} \theta} d\theta$$ which is equal to $$\int 4\cos^{2} \theta d\theta$$. now use the identity $$2\cos^{2}\theta = 1 + \cos(2\theta)$$

I've already got that one, but knowing how to do it that way would be good. I understand that identity, but can't see how that will help me..

Edit: Wait gimme a min, i think i see it..

Edit 2: I Get $$2\theta + \sin 2\theta$$ Then what do i do..

 

[murshid_islam]

And now for something a little different;

$$I = \int x^2\sin x\;dx$$

does the answer come out to be $$-x^2\cos^{2}x + 2x\sin x + 2\cos x + C$$?

 

[Hootenanny]

does the answer come out to be $$-x^2\cos^{2}x + 2x\sin x + 2\cos x + C$$?

I did integration by parts a few times, I got $$I=\frac{\sin^2 x}{2} - x^2 \cos x$$ :D I think that's right.

Note: This ones for x^2 sin x dx.

Neither is correct, but murshid is closer than Gib.

 

[Gib Z]

Umm Ok Ok, Problem 1. I can't see what to do with murshid_islams method, and When I found the derivative of my solution to I, for x^2 sin x dx, I get (2x + cos x)sin x, that doesn't work..

Edit: Ok i was wrong lol, Ill try it again

 

[murshid_islam]

Neither is correct, but murshid is closer than Gib.

lol. the answer i came up with is
$$-x^2\cos x + 2x\sin x + 2\cos x + C$$
i think this is correct.

I Get $$2\theta + \sin 2\theta$$ Then what do i do..

you used the substitution $$x = \frac{1}{2}\tan\theta$$
from this, you get, $$\tan\theta = 2x$$ and from this, you get $$\theta = \arctan(2x)$$ and $$\sin\theta = \frac{2x}{\sqrt{4x^2+1}}$$ and $$\cos\theta = \frac{1}{\sqrt{4x^2+1}}$$.

and you also know that $$\sin 2\theta = 2\sin\theta\cos\theta = \frac{4x}{4x^2+1}$$

so your answer will be
$$2\theta + \sin2\theta = 2\arctan(2x) + \frac{4x}{4x^2+1}$$

 

[Hootenanny]

lol. the answer i came up with is
$$-x^2\cos x + 2x\sin x + 2\cos x + C$$
i think this is correct.

This is indeed correct.

 

[Gib Z]

Ok, I made a few mistakes the first time, I got $$I = x^2 + 2x \sin x + \cos x + C$$, but when I derive it it doesn't work again :'(

EDIT: AWW KILL ME! Sorry Guys, I got the same answer and Murshid, Its just that during my working I wrote down $$x^2 -\cos x$$...It was ment to me the multiplication of these 2, or -x^2 (cos x), but I took it as x^2 minus cos x...:'( sorry guys